LeetCode——3Sum

1. Question

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]
]

2. Solution

1. 先排序，然后对于每一个nums[i],在其后找到两个数字num[j]和num[k]，j<k，使得三者和为0。

2. 去掉重复的组合，如果满足三者为0，那么j需要后移到一个不同的数，k也需要前移到一个不同的数。

3. 重复的nums[i]也需要去掉，因为nums[i]和nums[i+1] 都是和其后的进行组合，如果两个一样的话，就重复了，需要去掉。

3. Code

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        if (nums.size() <= 2)
            return vector<vector<int>>();
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        for (int i = 0; i < nums.size() - 2; i++) {
            int start = i + 1;
            int end = nums.size() - 1;
            while (start < end) {
                int tmp = nums[start] + nums[end];
                int target = -nums[i];
                if (tmp == target) {
                    vector<int> ele;
                    ele.push_back(nums[i]);
                    ele.push_back(nums[start]);
                    ele.push_back(nums[end]);
                    res.push_back(ele);
                    
                    // 去掉start重复的
                    while (start < end && nums[start] == ele[1])
                        start++;
                    // 去掉end重复的
                    while (end > start && nums[end] == ele[2])
                        end--;
                } else if (tmp > target)
                    end--;
                else
                    start++;
            }
            // 去掉nums[i]重复的
            while (i + 1 < nums.size() - 2 && nums[i + 1] == nums[i])
                i++;
        }
        return res;
    }
};