LintCode: 3 Sum

C++

把3个数求和，转变为2个数求和

1. 把数组排序

2. 注意过滤重复值

3. 从前到后遍历，游标i

4. 从后边数中找start ＋ end ＝ -arr[i]的2 sum

5. start ＋ end < -arr[i], start++

6. start + end > -arr[i], end--

7. start + end = -arr[i], insert <i, start, end> into result vecotr

class Solution {
public:    
    /**
     * @param numbers : Give an array numbers of n integer
     * @return : Find all unique triplets in the array which gives the sum of zero.
     */
    vector<vector<int> > threeSum(vector<int> &nums) {
        // write your code here
        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;
    }
};