Commit 7e0ea2c

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weiy

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count of smaller numbers after self hard

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	`1`	`+"""`
	`2`	`+You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].`
	`3`	`+`
	`4`	`+Example:`
	`5`	`+`
	`6`	`+Input: [5,2,6,1]`
	`7`	`+Output: [2,1,1,0]`
	`8`	`+Explanation:`
	`9`	`+To the right of 5 there are 2 smaller elements (2 and 1).`
	`10`	`+To the right of 2 there is only 1 smaller element (1).`
	`11`	`+To the right of 6 there is 1 smaller element (1).`
	`12`	`+To the right of 1 there is 0 smaller element.`
	`13`	`+`
	`14`	`+思路:`
	`15`	`+二分。`
	`16`	`+`
	`17`	`+瓶颈依然在于插入列表中的时候需要的时间复杂度为 O(n)。`
	`18`	`+`
	`19`	`+beat`
	`20`	`+82%`
	`21`	`+`
	`22`	`+测试地址:`
	`23`	`+https://leetcode.com/problems/count-of-smaller-numbers-after-self/description/`
	`24`	`+`
	`25`	`+"""`
	`26`	`+import bisect`
	`27`	`+`
	`28`	`+class Solution(object):`
	`29`	`+ def countSmaller(self, nums):`
	`30`	`+ """`
	`31`	`+ :type nums: List[int]`
	`32`	`+ :rtype: List[int]`
	`33`	`+ """`
	`34`	`+`
	`35`	`+ x = []`
	`36`	`+ result = []`
	`37`	`+`
	`38`	`+ for i in nums[::-1]:`
	`39`	`+ result.append(bisect.bisect_left(x, i))`
	`40`	`+ bisect.insort(x,i)`
	`41`	`+`
	`42`	`+ return result[::-1]`

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