15. 3Sum

发表于 | 分类于 | |

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: The solution set must not contain duplicate triplets.

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> > For example, given array S = [-1, 0, 1, 2, -1, -4],
> >
> > A solution set is:
> > [
> >   [-1, 0, 1],
> >   [-1, -1, 2]
> > ]
> >
>
>
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class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int len = nums.size();
        sort(nums.begin(), nums.end());
        int findNum;
        vector<vector<int>>::iterator iter;
        vector<vector<int>> res;
        for (int i = 0; i < len - 2; ++i)
        {
            if (nums[i] > 0)
                return res;
            for (int j = i + 1; j < len - 1; ++j)
            {
                findNum = 0 - nums[i] - nums[j];
                if (findNum < 0)
                    break;
                if (binary_search(nums.begin() + j + 1, nums.end(), findNum))
                {
                    vector<int> v;
                    v.push_back(nums[i]);
                    v.push_back(nums[j]);
                    v.push_back(findNum);
                    iter = std::find(res.begin(), res.end(), v);
                    if (iter == res.end())
                        res.push_back(v);
                }
            }
        }
        return res;
    }
};

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//
// Created by chaopengz on 2017/9/27.
//
#include "head.h"

class Solution {
public:
    vector<vector<int>> threeSum(vector<int> &nums)
    {
        int len = nums.size();
        int j, k, sum;
        vector<vector<int>> res;


        sort(nums.begin(), nums.end());

        for (int i = 0; i < len - 2; ++i)
        {
            if (i == 0 || nums[i] != nums[i - 1])//a = nums[i]取值不重复
            {
                j = i + 1;
                k = len - 1;
                sum = 0 - nums[i];
                while (j < k)
                {
                    if (nums[j] + nums[k] == sum)
                    {
                        vector<int> v;
                        v.push_back(nums[i]);
                        v.push_back(nums[j++]);
                        v.push_back(nums[k--]);
                        res.push_back(v);
                        while (j < k && nums[j] == nums[j - 1]) j++;
                        while (j < k && nums[k] == nums[k + 1]) k--;
                    } else if (nums[j] + nums[k] > sum)
                    {
                        k--;
                    } else
                    {
                        j++;
                    }
                }
            }
        }
        return res;
    }
};