leetcode 315. Count of Smaller Numbers After Self

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.

题目 返回每个位置右侧比当前位置数字小的数量。

从右向左,插入即可。代码如下:

class Solution(object):
    def countSmaller(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        result = []
        sortedList = []
        for num in nums[::-1]:
            position=bisect.bisect_left(sortedList, num)
            result.append(position)
            sortedList.insert(position,num)
        return result[::-1]

 

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