LeetCode-15. 3Sum [C++]

2021/12/25 17:10:15

本文主要是介绍LeetCode-15. 3Sum [C++],对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

LeetCode-15. 3Sumicon-default.png?t=LA92https://leetcode.com/problems/3sum/

题目描述

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

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

解题思路 

【C++解法】

class Solution {
public:
    vector<vector<int>> threeSum(vector<int> &nums) {
        vector< vector<int>> res;
        if(nums.size()<3) return res;
        sort(nums.begin(),nums.end());
        if(nums.front()>0 || nums.back()<0) return res;
        for(int i=0; i<nums.size()-2; ++i) {
            auto target = 0-nums[i];
            if(nums[i]+nums[i+1]>target) break;
            if(nums[nums.size()-1]+nums[nums.size()-2]<target) continue;
            int front = i+1, back = nums.size()-1;
            while(front<back) {
                auto sum = nums[front]+nums[back];
                if(sum<target) front++;
                else if(sum>target) back--;
                else {
                    res.push_back(vector<int>{nums[i],nums[front++],nums[back--]});
                    while(front<back && nums[front-1]==nums[front]) front++;
                    while(front<back && nums[back+1]==nums[back]) back--;
                }   
            }
            while(i<nums.size()-2 && nums[i]==nums[i+1]) i++;
        }
        return res;
    }
};



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