【leetcode】1365. How Many Numbers Are Smaller Than the Current Number

题目如下：

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation: 
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). 
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1). 
For nums[3]=2 there exist one smaller number than it (1). 
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:

Input: nums = [6,5,4,8]
Output: [2,1,0,3]

Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0] 

Constraints:

• 2 <= nums.length <= 500
• 0 <= nums[i] <= 100

解题思路：排序后用二分查找定位每个元素所在的位置，即可求出比自己小的数字的数量。

代码如下：

class Solution(object):
    def smallerNumbersThanCurrent(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        sort_nums = sorted(nums)
        import bisect
        res = []
        for num in nums:
            inx = bisect.bisect_left(sort_nums,num)
            res.append(inx)
        return res