leetcode 1365. How Many Numbers Are Smaller Than the Current Number（python）-爱代码爱编程

描述

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]

Note:

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

解析

根据题意，只需要在 nums 中找出每个元素的比其小的元素的个数，然后组成列表即可。所以将 nums 升序排序为 tmp ，然后在遍历 nums 的时候只需要定位每个元素在 tmp 中第一次出现的索引追加到 res 中，遍历结束将得到的答案。

解答

class Solution(object):
    def smallerNumbersThanCurrent(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        tmp = sorted(nums)
        res = []
        for i in nums:
            res.append(tmp.index(i))
        return res

运行结果

Runtime: 48 ms, faster than 74.62% of Python online submissions for How Many Numbers Are Smaller Than the Current Number.
Memory Usage: 13.5 MB, less than 37.19% of Python online submissions for How Many Numbers Are Smaller Than the Current Number.

原题链接：https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/