Leetcode 1365. How Many Numbers Are Smaller Than the Current Number

Tyan

发布于 2021-07-08 16:22:47

30800

代码可运行

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代码可运行

文章作者：Tyan 博客：noahsnail.com | CSDN | 简书

1. Description

2. Solution

**解析：**Version 1是两两比较所有的数，简单直接。Version 2，由于所有的数都在[0, 100]之间，因此统计[0,100]之间的数字个数即可，比98小的数是[0,97]之间数字的个数之和。Version 3是Version 2的通用版，不用限制数字的大小。Version 4是进一步优化，只统计出现的数字数量。

Version 1

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        n = len(nums)
        result = [0 for _ in range(n)]
        for i in range(n):
            for j in range(i+1, n):
                if nums[i] > nums[j]:
                    result[i] += 1
                elif nums[i] < nums[j]:
                    result[j] += 1
        return result

Version 2

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        stat = {i: 0 for i in range(101)}
        for num in nums:
            stat[num] = stat.get(num, 0) + 1

        count = 0
        for i in range(101):
            pre = count
            count += stat[i]
            stat[i] = pre

        result = [stat[num] for num in nums]
        return result

Version 3

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        maximun = max(nums)
        stat = {i: 0 for i in range(maximun+1)}
        for num in nums:
            stat[num] = stat.get(num, 0) + 1

        count = 0
        for i in range(maximun+1):
            pre = count
            count += stat[i]
            stat[i] = pre

        result = [stat[num] for num in nums]
        return result

Version 4

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        order = sorted(set(nums))
        stat = {num: 0 for num in order}
        for num in nums:
            stat[num] = stat.get(num, 0) + 1

        count = 0
        for num in order:
            pre = count
            count += stat[num]
            stat[num] = pre

        result = [stat[num] for num in nums]
        return result