57. 3Sum【medium】

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.

Notice

Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)

The solution set must not contain duplicate triplets.

 

Example

For example, given array S = {-1 0 1 2 -1 -4}, A solution set is:

(-1, 0, 1)
(-1, -1, 2)

题意

给出一个有n个整数的数组S，在S中找到三个整数a, b, c，找到所有使得a + b + c = 0的三元组。

在三元组(a, b, c)，要求a <= b <= c。结果不能包含重复的三元组。

解法一：

 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) {
         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++;
                     end--;
                 }
             }
         }

         return result;
     }
 };

解法二：

 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> &numbers) {
         vector<vector<int>> ret;

         int n = numbers.size();
         sort(numbers.begin(), numbers.end());

         for (int i = ; i < n - ; ++i) {
             if (i !=  && numbers[i] == numbers[i-]) {
                 continue;
             }

             int sum = -numbers[i];
             int j = i + , k = n - ;

             while (j < k) {
                 int tmp = numbers[j] + numbers[k];
                 if (tmp == sum) {
                     vector<int> sol{numbers[i], numbers[j], numbers[k]};
                     ret.push_back(sol);
                     while (j < k && numbers[j] == numbers[j+]) {
                         j++;
                     }
                     while (j < k && numbers[k] == numbers[k-]) {
                         k--;
                     }
                     j++;
                     k--;
                 } else if (tmp > sum) {
                     k--;
                 } else {
                     j++;
                 }
             }
         }
         return ret;
     }
 };