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feat: add solutions to lc problem: No.2087
No.2087.Minimum Cost Homecoming of a Robot in a Grid
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‎solution/2000-2099/2085.Count Common Words With One Occurrence/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表计数**
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**方法一：哈希表**
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我们可以用两个哈希表分别统计两个字符串数组中每个字符串出现的次数，然后遍历其中一个哈希表，如果某个字符串在另一个哈希表中出现了一次，且在当前哈希表中也出现了一次，则答案加一。
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时间复杂度 $O(n + m)$，空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别是两个字符串数组的长度。
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<!-- tabs:start -->
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‎solution/2000-2099/2087.Minimum Cost Homecoming of a Robot in a Grid/README.md

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@@ -59,22 +59,122 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心**
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设机器人当前位置为 $(i, j)$，目标位置为 $(x, y)$。
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- 如果 $i \lt x$，则机器人往下移动，代价为 $rowCosts[i + 1] + rowCosts[i + 2] + \cdots + rowCosts[x]$。
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- 如果 $i \gt x$，则机器人往上移动，代价为 $rowCosts[x] + rowCosts[x + 1] + \cdots + rowCosts[i - 1]$。
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- 如果 $j \lt y$，则机器人往右移动，代价为 $colCosts[j + 1] + colCosts[j + 2] + \cdots + colCosts[y]$。
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- 如果 $j \gt y$，则机器人往左移动，代价为 $colCosts[y] + colCosts[y + 1] + \cdots + colCosts[j - 1]$。
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时间复杂度 $O(m + n)$，空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别为 $rowCosts$ 和 $colCosts$ 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minCost(self, startPos: List[int], homePos: List[int], rowCosts: List[int], colCosts: List[int]) -> int:
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i, j = startPos
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x, y = homePos
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ans = 0
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if i < x:
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ans += sum(rowCosts[i + 1: x + 1])
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else:
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ans += sum(rowCosts[x: i])
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if j < y:
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ans += sum(colCosts[j + 1: y + 1])
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else:
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ans += sum(colCosts[y: j])
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int minCost(int[] startPos, int[] homePos, int[] rowCosts, int[] colCosts) {
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int i = startPos[0], j = startPos[1];
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int x = homePos[0], y = homePos[1];
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int ans = 0;
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if (i < x) {
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for (int k = i + 1; k <= x; ++k) {
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ans += rowCosts[k];
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}
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} else {
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for (int k = x; k < i; ++k) {
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ans += rowCosts[k];
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}
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}
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if (j < y) {
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for (int k = j + 1; k <= y; ++k) {
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ans += colCosts[k];
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}
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} else {
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for (int k = y; k < j; ++k) {
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ans += colCosts[k];
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minCost(vector<int>& startPos, vector<int>& homePos, vector<int>& rowCosts, vector<int>& colCosts) {
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int i = startPos[0], j = startPos[1];
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int x = homePos[0], y = homePos[1];
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int ans = 0;
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if (i < x) {
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ans += accumulate(rowCosts.begin() + i + 1, rowCosts.begin() + x + 1, 0);
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} else {
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ans += accumulate(rowCosts.begin() + x, rowCosts.begin() + i, 0);
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}
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if (j < y) {
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ans += accumulate(colCosts.begin() + j + 1, colCosts.begin() + y + 1, 0);
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} else {
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ans += accumulate(colCosts.begin() + y, colCosts.begin() + j, 0);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minCost(startPos []int, homePos []int, rowCosts []int, colCosts []int) (ans int) {
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i, j := startPos[0], startPos[1]
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x, y := homePos[0], homePos[1]
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if i < x {
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ans += sum(rowCosts, i+1, x+1)
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} else {
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ans += sum(rowCosts, x, i)
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}
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if j < y {
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ans += sum(colCosts, j+1, y+1)
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} else {
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ans += sum(colCosts, y, j)
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}
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return
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}
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func sum(nums []int, i, j int) (s int) {
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for k := i; k < j; k++ {
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s += nums[k]
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}
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return
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}
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```
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### **...**

‎solution/2000-2099/2087.Minimum Cost Homecoming of a Robot in a Grid/README_EN.md

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@@ -58,13 +58,102 @@ The total cost is 3 + 2 + 6 + 7 = 18</pre>
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### **Python3**
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```python
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class Solution:
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def minCost(self, startPos: List[int], homePos: List[int], rowCosts: List[int], colCosts: List[int]) -> int:
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i, j = startPos
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x, y = homePos
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ans = 0
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if i < x:
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ans += sum(rowCosts[i + 1: x + 1])
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else:
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ans += sum(rowCosts[x: i])
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if j < y:
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ans += sum(colCosts[j + 1: y + 1])
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else:
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ans += sum(colCosts[y: j])
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int minCost(int[] startPos, int[] homePos, int[] rowCosts, int[] colCosts) {
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int i = startPos[0], j = startPos[1];
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int x = homePos[0], y = homePos[1];
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int ans = 0;
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if (i < x) {
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for (int k = i + 1; k <= x; ++k) {
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ans += rowCosts[k];
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}
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} else {
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for (int k = x; k < i; ++k) {
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ans += rowCosts[k];
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}
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}
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if (j < y) {
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for (int k = j + 1; k <= y; ++k) {
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ans += colCosts[k];
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}
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} else {
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for (int k = y; k < j; ++k) {
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ans += colCosts[k];
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minCost(vector<int>& startPos, vector<int>& homePos, vector<int>& rowCosts, vector<int>& colCosts) {
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int i = startPos[0], j = startPos[1];
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int x = homePos[0], y = homePos[1];
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int ans = 0;
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if (i < x) {
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ans += accumulate(rowCosts.begin() + i + 1, rowCosts.begin() + x + 1, 0);
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} else {
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ans += accumulate(rowCosts.begin() + x, rowCosts.begin() + i, 0);
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}
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if (j < y) {
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ans += accumulate(colCosts.begin() + j + 1, colCosts.begin() + y + 1, 0);
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} else {
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ans += accumulate(colCosts.begin() + y, colCosts.begin() + j, 0);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func minCost(startPos []int, homePos []int, rowCosts []int, colCosts []int) (ans int) {
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i, j := startPos[0], startPos[1]
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x, y := homePos[0], homePos[1]
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if i < x {
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ans += sum(rowCosts, i+1, x+1)
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} else {
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ans += sum(rowCosts, x, i)
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}
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if j < y {
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ans += sum(colCosts, j+1, y+1)
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} else {
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ans += sum(colCosts, y, j)
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}
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return
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}
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func sum(nums []int, i, j int) (s int) {
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for k := i; k < j; k++ {
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s += nums[k]
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}
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return
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}
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```
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### **...**

‎solution/2000-2099/2087.Minimum Cost Homecoming of a Robot in a Grid/Solution.cpp

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class Solution {
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public:
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int minCost(vector<int>& startPos, vector<int>& homePos, vector<int>& rowCosts, vector<int>& colCosts) {
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int i = startPos[0], j = startPos[1];
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int x = homePos[0], y = homePos[1];
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int ans = 0;
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if (i < x) {
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ans += accumulate(rowCosts.begin() + i + 1, rowCosts.begin() + x + 1, 0);
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} else {
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ans += accumulate(rowCosts.begin() + x, rowCosts.begin() + i, 0);
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}
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if (j < y) {
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ans += accumulate(colCosts.begin() + j + 1, colCosts.begin() + y + 1, 0);
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} else {
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ans += accumulate(colCosts.begin() + y, colCosts.begin() + j, 0);
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}
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return ans;
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}
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};

‎solution/2000-2099/2087.Minimum Cost Homecoming of a Robot in a Grid/Solution.go

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func minCost(startPos []int, homePos []int, rowCosts []int, colCosts []int) (ans int) {
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i, j := startPos[0], startPos[1]
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x, y := homePos[0], homePos[1]
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if i < x {
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ans += sum(rowCosts, i+1, x+1)
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} else {
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ans += sum(rowCosts, x, i)
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}
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if j < y {
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ans += sum(colCosts, j+1, y+1)
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} else {
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ans += sum(colCosts, y, j)
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}
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return
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}
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func sum(nums []int, i, j int) (s int) {
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for k := i; k < j; k++ {
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s += nums[k]
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}
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return
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}

‎solution/2000-2099/2087.Minimum Cost Homecoming of a Robot in a Grid/Solution.java

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class Solution {
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public int minCost(int[] startPos, int[] homePos, int[] rowCosts, int[] colCosts) {
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int i = startPos[0], j = startPos[1];
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int x = homePos[0], y = homePos[1];
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int ans = 0;
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if (i < x) {
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for (int k = i + 1; k <= x; ++k) {
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ans += rowCosts[k];
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}
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} else {
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for (int k = x; k < i; ++k) {
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ans += rowCosts[k];
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}
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}
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if (j < y) {
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for (int k = j + 1; k <= y; ++k) {
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ans += colCosts[k];
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}
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} else {
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for (int k = y; k < j; ++k) {
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ans += colCosts[k];
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}
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}
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return ans;
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}
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}

‎solution/2000-2099/2087.Minimum Cost Homecoming of a Robot in a Grid/Solution.py

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class Solution:
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def minCost(self, startPos: List[int], homePos: List[int], rowCosts: List[int], colCosts: List[int]) -> int:
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i, j = startPos
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x, y = homePos
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ans = 0
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if i < x:
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ans += sum(rowCosts[i + 1: x + 1])
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else:
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ans += sum(rowCosts[x: i])
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if j < y:
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ans += sum(colCosts[j + 1: y + 1])
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else:
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ans += sum(colCosts[y: j])
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return ans

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