25 Reverse Nodes in k-Group

Author: int32bit

README.md

@@ -19,6 +19,7 @@
 + [22 Generate Parentheses(递归、搜索)](algorithms/GenerateParentheses)
 + [23 Merge k Sorted Lists(堆)](algorithms/MergekSortedLists)
 + [24 Swap Nodes in Pairs(链表操作)](algorithms/SwapNodesinPairs)
++ [25 Reverse Nodes in k-Group(链表操作)](algorithms/ReverseNodesink-Group)
 + [26 Remove Duplicates from Sorted Array](algorithms/RemoveDuplicatesfromSortedArray)
 + [27 Remove Element](algorithms/RemoveElement)
 + [29 Divide Two Integers(位运算、最小负数绝对值、溢出检测)](algorithms/DivideTwoIntegers)

algorithms/ReverseNodesink-Group/README.md

@@ -0,0 +1,110 @@
+## Reverse Nodes in k-Group
+
+Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
+
+If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
+
+You may not alter the values in the nodes, only nodes itself may be changed.
+
+Only constant memory is allowed.
+
+For example,
+
+```
+Given this linked list: 1->2->3->4->5
+
+For k = 2, you should return: 2->1->4->3->5
+
+For k = 3, you should return: 3->2->1->4->5 
+```
+
+## Solution
+
+首先定义一个reverse函数，接收两个指针，分别为begin和end，表示reverse的开始和结束，同时由于end的next指针会丢失，因此应该预先保存并返回
+
+```cpp
+ListNode *reverse(ListNode *begin, ListNode *end) {
+	assert(begin && end);
+	ListNode *p = begin;
+	ListNode *prev = nullptr;
+	ListNode *endFlag = end->next;
+	while (p != endFlag) { // 条件不能写成 p != end->next, 因为end->next reverse时会修改
+		ListNode *q = p->next; // 保存next指针
+		p->next = prev;
+		prev = p;
+		p = q;
+	}
+	return p;
+}
+```
+
+然后分组，思路是每次走k步，找到需要reverse的开始和结束节点，同时注意保存之前已经处理好的尾部节点prev
+
+```cpp
+while (begin) {
+	int K = k - 1; // 包括begin节点，因此后面还需要k-1个节点
+	ListNode *p = begin;
+	while (K && p->next) {
+		end = p->next;
+		p = p->next;
+		K--;
+	}
+	if (K) { // 不足k，不需要reverse，但注意需要更新尾部节点
+		prev->next = begin; // 记得最后更新prev指针
+		return head;
+	}
+	ListNode *next = reverse(begin, end);
+	prev->next = end; // 更新已经处理完毕的尾部节点
+	prev = begin; // 更新处理完毕的尾部节点指针
+	begin = next; // begin从下一个节点开始
+}
+```
+
+完整代码:
+
+```cpp
+ListNode *reverseKGroup(ListNode *head, int k) {
+	if (head == nullptr || k <= 1)
+		return head;
+	ListNode *begin = head, *end = nullptr;
+	ListNode *prev = nullptr;
+	/* First reverse */
+	if (begin) { // 第一次操作，需要更新head和prev指针
+		int K = k - 1;
+		ListNode *p = begin;
+		while (K && p->next) {
+			end = p->next;
+			p = p->next;
+			K--;
+		}
+		if (K)
+			return head;
+		ListNode *next = reverse(begin, end);
+		prev = begin;
+		head = end;
+		begin = next;
+	}
+	while (begin) {
+		int K = k - 1;
+		ListNode *p = begin;
+		while (K && p->next) {
+			end = p->next;
+			p = p->next;
+			K--;
+		}
+		if (K) {
+			prev->next = begin; // 记得最后更新prev指针
+			return head;
+		}
+		ListNode *next = reverse(begin, end);
+		prev->next = end;
+		prev = begin;
+		begin = next;
+	}
+	return head;
+}
+```
+
+## 得瑟
+
+居然一次AC，:)

algorithms/ReverseNodesink-Group/list.cpp

@@ -0,0 +1,128 @@
+#include <iostream>
+#include <cstdio>
+#include <vector>
+#include <algorithm>
+#include <string>
+using namespace std;
+#include <stdlib.h>
+#include <cassert>
+struct ListNode {
+	int val;
+	ListNode *next;
+	ListNode(int x): val(x), next(nullptr) {}
+};
+class Solution {
+	public:
+		ListNode *reverseKGroup(ListNode *head, int k) {
+			if (head == nullptr || k <= 1)
+				return head;
+			ListNode *begin = head, *end = nullptr;
+			ListNode *prev = nullptr;
+			/* First reverse */
+			/* 第一次操作，主要需要更新head和prev指针 */
+			if (begin) {
+				int K = k - 1;
+				ListNode *p = begin;
+				while (K && p->next) {
+					end = p->next;
+					p = p->next;
+					K--;
+				}
+				if (K)
+					return head;
+				ListNode *next = reverse(begin, end);
+				prev = begin;
+				head = end;
+				begin = next;
+			}
+			while (begin) {
+				int K = k - 1;
+				ListNode *p = begin;
+				while (K && p->next) {
+					end = p->next;
+					p = p->next;
+					K--;
+				}
+				if (K) { // 不足k个节点，不需要reverse，但需要更新处理完毕的尾部节点
+					prev->next = begin; // 记得最后更新prev指针
+					return head;
+				}
+				ListNode *next = reverse(begin, end);
+				prev->next = end;
+				prev = begin;
+				begin = next;
+			}
+			return head;
+		}
+	private:
+		ListNode *reverse(ListNode *begin, ListNode *end) {
+			assert(begin && end);
+			ListNode *p = begin;
+			ListNode *prev = nullptr;
+			ListNode *endFlag = end->next;
+			while (p != endFlag) { // 条件不能写成 p != end->next, 因为end->next reverse时会修改
+				ListNode *q = p->next;
+				p->next = prev;
+				prev = p;
+				p = q;
+			}
+			return p;
+		}
+
+};
+int getLength(ListNode *head)
+{
+	int len = 0;
+	ListNode *p = head;
+	while (p) {
+		++len;
+		p = p->next;
+	}
+	return len;
+}
+void print(ListNode *head)
+{
+	if (head == nullptr) {
+		printf("NULL\n");
+		return;
+	}
+	struct ListNode *p = head;
+	while (p) {
+		printf("%d ", p->val);
+		p = p->next;
+	}
+	printf("\n");
+}
+ListNode * mk_list(ListNode **ha, int a[], int n)
+{
+	if (n < 1)
+		return nullptr;
+	ListNode *p = new ListNode(a[0]);
+	*ha = p;
+	for (int i = 1; i < n; ++i) {
+		ListNode *q = new ListNode(a[i]);
+		p->next = q;
+		p = q;
+	}
+	return p;
+}
+void free_list(struct ListNode *head)
+{
+	struct ListNode *p = head;
+	while (p) {
+		struct ListNode *q = p->next;
+		delete p;
+		p = q;
+	}
+}
+int main(int argc, char **argv)
+{
+	Solution solution;
+	struct ListNode *head = NULL;
+	int a[] = {1, 2, 3, 4, 5, 6};
+	mk_list(&head, a, 6);
+	print(head);
+	head = solution.reverseKGroup(head, 3);
+	print(head);
+	return 0;
+}