Leetcode--3Sum

Problem Description:

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},A solution set is:(-1, 0, 1)(-1, -1, 2)

分析：题目要求把全部可能的集合都找出来。因此想到的就是首先将数组排序，然后利用两重循环依次选出两个数字a和b。然后在剩下的数字中查找是否存在c，详细实现用到了upper_bound函数找到比当前数大的第一个数，去掉反复循环的情况。然后用find函数查找c是否存在，存在则将三个数记录下来。

详细代码例如以下：

class Solution {
public:vector<vector<int> > threeSum(vector<int> &num) {vector<vector<int> > results;if(num.size()<3) return results;vector<int> subset;sort(num.begin(),num.end());vector<int>::iterator p=num.begin(),q,flag;while(p<(num.end()-2)){q=p+1;while(q<num.end()-1){int tag=0-*p-*q;if(find(q+1,num.end(),tag)!=num.end()){flag=find(q+1,num.end(),tag);subset.push_back(*p);subset.push_back(*q);subset.push_back(*flag);results.push_back(subset);}subset.clear();q=upper_bound(q,num.end()-1,*q);}p=upper_bound(p,num.end()-2,*p);}return results;}
};

转载于:https://www.cnblogs.com/liguangsunls/p/6880124.html

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