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 = ; i < nums.size(); i++) {
             if (i >  && nums[i] == nums[i - ]) {
                 continue;
             }
             // two sum;
             int start = i + , end = nums.size() - ;
             int target = -nums[i];
             while (start < end) {
                 if (start > i +  && nums[start - ] == 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;
     }
 };