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
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
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 != i
且 nums[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
最怕一生碌碌无为,还说平凡难能可贵。