LeetCode题解 —— 15. 3Sum

题目描述

Given an array nums of n integers, are there elements a, b, c 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]
]

解题思想

如果把等式做个移项，a + b + c = 0 -> a + b = -c，问题就转化为，寻找两个变量a和b，使其之和为-c，又回到了两数之和的问题。

先对数组排序，采用双指针的思想，从首尾往中间扫描。因为题目提到了可能出现重复解的问题，所以要注意去重：
* 双指针扫描的去重
* 外层循环的去重

解题代码

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> result;
        for(int i = 0; i < nums.size(); i++){
            int target = -nums[i];
            int start = i + 1;
            int end = nums.size() - 1;
            while(start < end){
                if(nums[start] + nums[end] < target)
                    start++;
                else if(nums[start] + nums[end] > target)
                    end--;
                else{
                    vector<int> temp;
                    temp.push_back(nums[i]);
                    temp.push_back(nums[start]);
                    temp.push_back(nums[end]);
                    result.push_back(temp);
                    start++;
                    end--;
                    while(start < end && nums[start] == nums[start - 1])
                        start++;
                    while(start < end && nums[end] == nums[end + 1])
                        end--;
                }
            }
            while(i + 1 < nums.size() && nums[i + 1] == nums[i])
                i++;
        }
        return result;
    }
};

int main()
{
    Solution solution;
    vector<int> nums = {-1, 0, 1, 2, -1, -4};
    vector<vector<int>> result = solution.threeSum(nums);
    for(auto re : result)
        cout << '[' << re[0] << ',' << re[1] << ',' << re[2] << ']' << endl;
    return 0;
}