15. 3Sum

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.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:

[

[-1, 0, 1],

[-1, -1, 2]

]

解析

此題的結(jié)題思想其實(shí)和TwoSum一樣，當(dāng)我們確定了第一個(gè)數(shù)字后，第二個(gè)和第三個(gè)數(shù)字就可以用TwoSum一樣的解法求得。 唯一的區(qū)別是TwoSum這題只有唯一解，而 3Sum這題可能有多個(gè)相同的解，我們需要考慮怎么移除相同的解。

因?yàn)榉祷氐氖蔷唧w的三個(gè)數(shù)字，而不是數(shù)字的下標(biāo)，所以可以考慮先將輸入的序列進(jìn)行排序。得到排序后的序列num，長(zhǎng)度為n。

vector<vector<int> > threeSum(vector<int> &num) {    vector<vector<int> > res;    std::sort(num.begin(), num.end());    for (int i = 0; i < num.size(); i++) {        int target = -num[i];        int front = i + 1;        int back = num.size() - 1;        while (front < back) {            int sum = num[front] + num[back];            // Finding answer which start from number num[i]            if (sum < target)                front++;            else if (sum > target)                back--;            else {                vector<int> triplet(3, 0);                triplet[0] = num[i];                triplet[1] = num[front];                triplet[2] = num[back];                res.push_back(triplet);                // PRocessing duplicates of Number 2                // Rolling the front pointer to the next different number forwards                while (front < back && num[front] == triplet[1]) front++;                // Processing duplicates of Number 3                // Rolling the back pointer to the next different number backwards                while (front < back && num[back] == triplet[2]) rear--;            }        }        // Processing duplicates of Number 1        while (i + 1 < num.size() && num[i + 1] == num[i])             i++;    }    return res;}
github：https://github.com/Subenle/LeetCode-in-Cpp/blob/master/015.md