leetcode_three_sum

发表于

难度：medium
本题可以看作是2sum的一种延伸，固定一个元素，对其余进行2sum。需要注意的是去除掉重复项和越界问题。代码如下。

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class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &nums) {
        vector<vector<int> > result;
        
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            // two sum;
            int start = i + 1, end = nums.size() - 1;
            int target = -nums[i];
            while (start < end) {
                if (start > i + 1 && nums[start - 1] == nums[start]) {
                    start++;
                    continue;
                }
                if (nums[start] + nums[end] < target) {
                    start++;
                } else if (nums[start] + nums[end] > target) {
                    end--;
                } else {
                    vector<int> triple;
                    triple.push_back(nums[i]);
                    triple.push_back(nums[start]);
                    triple.push_back(nums[end]);
                    result.push_back(triple);
                    start++;
                }
            }
        }
        
        return result;
    }
};

结果：46 ms，超过67.82%