LeetCode 15 3Sum（3个数的和）

翻译

给定一个有n个整数的数组S，是否存在三个元素a,b,c使得a+b+c=0?找出该数组中所有不重复的3个数，它们的和为0。备注：这三个元素必须是从小到大进行排序。结果中不能有重复的3个数。例如，给定数组S={-1 0 1 2 -1 4}，一个结果集为：(-1, 0, 1)(-1, -1, 2)

原文

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)

经典方法，可惜我并没有想到这样写……

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        sort(nums.begin(), nums.end());        vector<vector<int>> result;        int len = nums.size();              for (int current = 0; current < len - 2&&nums[current]<=0;current++)        {            int front = current + 1, back = len - 1;            while (front < back)            {                if (nums[current] + nums[front] + nums[back] < 0)                    front++;                else if (nums[current] + nums[front] + nums[back] > 0)                    back--;                else                {                    vector<int> v(3);                        v.push_back(nums[current]);                        v.push_back(nums[front]);                        v.push_back(nums[back]);                        result.push_back(v);                        v.clear();                    do {                        front++;                    } while (front < back&&nums[front - 1] == nums[front]);                    do {                        back--;                    } while (front < back&&nums[back + 1] == nums[back]);                }            }                                while (current < len - 2 && nums[current + 1] == nums[current])                current++;        }                                          return result;    }};

继续努力……

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