Time: 13 ms (47.95%), Space: 39 MB (73.10%) - LeetHub

Author: yeonjiyeon

1365-how-many-numbers-are-smaller-than-the-current-number/1365-how-many-numbers-are-smaller-than-the-current-number.java

@@ -0,0 +1,25 @@
+/*
+I:int[]
+O:int[]
+C:
+E:
+Solution
+1. nums를 돌면서 첫번째부터 차례대로 해당 값과 나머지 값들을 비교한다. 
+Time:O(n^2)
+Space:
+*/
+class Solution {
+    public int[] smallerNumbersThanCurrent(int[] nums) {
+        int[] res = new int[nums.length];
+        for(int i = 0; i < nums.length; i++){
+            int count = 0;
+            for(int j = 0; j < nums.length; j++){
+                if(nums[i]>nums[j]){
+                    count++;
+                }
+                res[i] = count;
+            }
+        }
+        return res;
+    }
+}