[Leetcode] (python): 015-3sum

Source of the topic:

https://leetcode.com/problems/3sum/

Test Instructions Analysis:

This topic is input an array of nums. Find out all 3 numbers so that the sum of the 3 numbers is 0.1. The 3 digits of the output are sorted by small to large, and the combination of 2.3 numbers is not repeated. For example, enter [ -1,0,1,2,-1,-4], which should return [[ -1,0,1],[-1,-1,2]].

Topic Ideas:

If the direct violence is resolved, then the time complexity is (O (n^3)). This will be tle.

Seeing this topic, I recalled the first question of Leetcode. The first question is to give an array and a target to find out the two numbers of the array so that the two numbers equals target. In the first question, I mentioned the "clamping theorem". This theorem can be used here. First , we'll sort the input array nums. then , from the beginning, take a number, nums[i], in nums[i+1:] with the clamping theorem to find Num[j],nums[k] (j<k) make their and for 0-nums[i]. Then [Nums[i],nums[j],nums[k]] Append to the answer array. Because there will be multiple combinations so nums[i] + nums[j] + nums[k] = 0, so at the time of comparison, if NUMS[J] + Nums[k] < 0-nums[i] time, j + = 1; if NUMS[J] + nums[k] > 0-nums[i] time, k-= 1; if NUMS[J] + nums[k] = = 0-nums[i], always j + = 1,k-= 1 until nums[j]! = Nums[j-1] and nums[k]! = nums[k + 1]. Note that in order to avoid duplicate combinations, i + should always be added to nums[i]! = nums[i-1].

In this method, the time complexity of sorting is (O (N*log (n))), the time complexity of the clamping theorem (o (n)), the complexity of the FETCH (O (n)), the total time complexity (O (N*log (n)) + O (n) *o (n)) = O (n^2). so the complexity of Time is O (n^2).

Code (Python):

1 classsolution (object):2     defthreesum (Self, nums):3         """4 : Type Nums:list[int]5 : Rtype:list[list[int]]6         """7Size =Len (nums)8Ans = []9         ifSize <= 2:Ten             returnans One Nums.sort () Ai =0 -          whileI < size-2: -TMP = 0-Nums[i] thej = i + 1 -K = size-1 -              whileJ <K: -                 ifNUMS[J] + Nums[k] <tmp: +J + = 1 -                 elifNUMS[J] + nums[k] >tmp: +K-= 1 A                 Else: at ans.append ([nums[i],nums[j],nums[k]]) -J + = 1 -K-= 1 -                      whileJ <K: -                         ifNUMS[J]! = nums[j-1]: -                              Break in                         ifNUMS[K]! = nums[k + 1]: -                              Break toJ + = 1 +K-= 1 -i + = 1 the              whileI < size-2: *                 ifNums[i]! = nums[i-1]: $                      BreakPanax Notoginsengi + = 1 -         returnAns

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Reprint Please specify source: http://www.cnblogs.com/chruny/p/4820473.html

[Leetcode] (python): 015-3sum