LeetCode | 1365. How Many Numbers Are Smaller Than the Current Number有多少小于当前数字的数字【Python】

LeetCode 1365. How Many Numbers Are Smaller Than the Current Number有多少小于当前数字的数字【Easy】【Python】【暴力】

Problem

LeetCode

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

问题

力扣

给你一个数组 nums,对于其中每个元素 nums[i],请你统计数组中比它小的所有数字的数目。

换而言之,对于每个 nums[i] 你必须计算出有效的 j 的数量,其中 j 满足 j != inums[j] < nums[i]

以数组形式返回答案。

示例 1:

输入:nums = [8,1,2,2,3]
输出:[4,0,1,1,3]
解释: 
对于 nums[0]=8 存在四个比它小的数字:(1,2,2 和 3)。 
对于 nums[1]=1 不存在比它小的数字。
对于 nums[2]=2 存在一个比它小的数字:(1)。 
对于 nums[3]=2 存在一个比它小的数字:(1)。 
对于 nums[4]=3 存在三个比它小的数字:(1,2 和 2)。

示例 2:

输入:nums = [6,5,4,8]
输出:[2,1,0,3]

示例 3:

输入:nums = [7,7,7,7]
输出:[0,0,0,0]

提示:

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

思路

暴力

两行 for 循环暴力一下就过了。

时间复杂度: O(n^2)
空间复杂度: O(n)

Python3代码
class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = []
        for i in range(n):
            cnt = 0
            for j in range(n):
                if nums[j] < nums[i]:
                    cnt += 1
            ans.append(cnt)
        return ans

代码地址

GitHub链接

posted @ 2020-03-01 15:36  Wonz  阅读(280)  评论(0编辑  收藏  举报