leetcode134:3sum

题目描述

给出一个有n个元素的数组S，S中是否有元素a,b,c满足a+b+c=0？找出数组S中所有满足条件的三元组。

注意：

1. 三元组（a、b、c）中的元素必须按非降序排列。（即a≤b≤c）
2. 解集中不能包含重复的三元组。

    例如，给定的数组 S = {-1 0 1 2 -1 -4},解集为(-1, 0, 1) (-1, -1, 2)

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)

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &nums) {
        vector<vector<int>> res;
        sort(nums.begin(),nums.end());
        for (int k=0;k<nums.size();++k){
            if (nums[k]>0)break;
            if (k>0 && nums[k]==nums[k-1]) continue;
            int target=0-nums[k];
            int i=k+1,j=nums.size()-1;
            while (i<j){
                if (nums[i]+nums[j]==target){
                    res.push_back({nums[k],nums[i],nums[j]});
                    while (i<j && nums[i]==nums[i+1]) ++i;
                    while (i<j && nums[j]==nums[j-1]) --j;
                    ++i;--j;
                    
                }else if (nums[i]+nums[j]<target)++i;
                   else --j;
            }
        }
        return res;
    }
};
class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        sort(num.begin(),num.end());
        vector <vector<int>> ans;
        for (int i=0;i<num.size();i++){
            if (i==0 || num[i]!=num[i-1]){
                int left=i+1,right=num.size()-1;
                while (left<right){
                    while (left<right && num[i]+num[left]+num[right]>0) right--;
                    if (left<right && num[i]+num[left]+num[right]==0){
                        vector<int> temp(3);
                        temp[0]=num[i];
                        temp[1]=num[left];
                        temp[2]=num[right];
                        ans.push_back(temp);
                        while (left<right && num[left]==temp[1]) left++;
                    }else {
                        left++;
                    }
                    }
                }
            }
        
return ans;
    }
};