1. 题目

3Sum

2.算法

• http://bangbingsyb.blogspot.jp/2014/11/leetcode-3sum.html

• 先排序，再左右夹逼：O(n^2)

O(n^2)=O(nlgn)+n*twopointers(n):O(n^2)

3.代码

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        if(nums.size()<2)  return res;
        sort(nums.begin(), nums.end());
        for (int k = 0; k < nums.size(); ++k) {
            if (nums[k] > 0) break;
            if (k > 0 && nums[k] == nums[k - 1]) continue;
            int target = 0 - nums[k];
            int i = k + 1, j = nums.size() - 1;
            while (i < j) {
                if (nums[i] + nums[j] == target) {
                    res.push_back({nums[k], nums[i], nums[j]});
                    while (i < j && nums[i] == nums[i + 1]) ++i;
                    while (i < j && nums[j] == nums[j - 1]) --j;
                    ++i; --j;
                } else if (nums[i] + nums[j] < target) ++i;
                else --j;
            }
        }
        return res;
    }
};