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 = 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++;
                    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 = 0; i < n - 2; ++i) {
            if (i != 0 && numbers[i] == numbers[i-1]) {
                continue;
            }
            
            int sum = -numbers[i];
            int j = i + 1, k = n - 1;
            
            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+1]) { 
                        j++;
                    }
                    while (j < k && numbers[k] == numbers[k-1]) {
                        k--;
                    }
                    j++; 
                    k--;
                } else if (tmp > sum) {
                    k--;
                } else {
                    j++;
                }
            }
        }
        return ret;
    }
};