count of smaller numbers after self hard

Author: weiy

Diff for: Array/CountOfSmallerNumbersAfterSelf.py

@@ -0,0 +1,42 @@
+"""
+You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].
+
+Example:
+
+Input: [5,2,6,1]
+Output: [2,1,1,0] 
+Explanation:
+To the right of 5 there are 2 smaller elements (2 and 1).
+To the right of 2 there is only 1 smaller element (1).
+To the right of 6 there is 1 smaller element (1).
+To the right of 1 there is 0 smaller element.
+
+思路：
+二分。
+
+瓶颈依然在于插入列表中的时候需要的时间复杂度为 O(n)。
+
+beat
+82%
+
+测试地址:
+https://leetcode.com/problems/count-of-smaller-numbers-after-self/description/
+
+"""
+import bisect
+
+class Solution(object):
+    def countSmaller(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        
+        x = []
+        result = []
+        
+        for i in nums[::-1]:
+            result.append(bisect.bisect_left(x, i))
+            bisect.insort(x,i)
+        
+        return result[::-1]