LeetCode 15. 三数之和 3Sum

先排序，然后固定一个值，使用双指针计算结果。

时间复杂度O(n^2)，空间复杂度O(1)

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int> > res;
        int length = nums.size();
        if (length < 3) return res;

        sort(nums.begin(), nums.end());
        for (int i = 0; i < length; ++i)
        {
            if (nums[i] > 0) break;  //最小值大于0
            if (i > 0 && nums[i] == nums[i - 1]) continue;  //处理重复值
            int left = i + 1, right = length - 1;
            while (left < right)
            {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum == 0)
                {
                    vector<int> temp = {nums[i], nums[left], nums[right]};
                    res.push_back(temp);
                    while (left < right && nums[left] == nums[left + 1]) ++left;
                    while (left < right && nums[right] == nums[right - 1]) --right;
                    ++left;
                    --right;
                }
                else if (sum > 0) --right;
                else ++left;
            }
        }
        return res;
    }
};