【Leetcode】15. 3Sum

题目描述

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]]

解题思路

双指针问题。
当固定住其中一个值时候，题目就和Leetcode1 2sum差不多了。
第二个数字从第一个数字的后一个开始找，而第三个数字则从最后一个数字开始找，通过移动控制这两个indexer的移动获取到正确的组合。
然后向后移动，更新第一个数字，再次查找其对应的所有剩下两个数字的组合。
为了防止重复，当每个数字被使用过之后，跳过与之相同的数字。

AC代码

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        vector<vector<int>> ans;        sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size(); ++i) {            // fix the first number and find the other two from left and right            int target = -nums[i];            int front = i + 1;            int back = nums.size() - 1;            while (front < back) {                int twoSum = nums[front] + nums[back];                if (twoSum > target)                    back--;                else if (twoSum < target)                    front++;                else {                    vector<int> triplets;                    triplets.push_back(nums[i]);                    triplets.push_back(nums[front]);                    triplets.push_back(nums[back]);                    ans.push_back(triplets);                    //get rid of duplicated number two and three                    while(front < back && nums[front] == triplets[1])                         front++;                    while(front < back && nums[back] == triplets[2])                        back--;                }            }            while(i + 1 < nums.size() && nums[i + 1] == nums[i])                i++;        }        return ans;    }};