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]
]

第一遍寫了下面的代碼，是錯的：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> answers;
        int n=0;
        if(nums.empty() || nums.capacity() < 3) return answers;
        for(int i=0; i < nums.capacity() - 2; i++)
            
 
for(int j = i + 1; j < nums.capacity() - 1; j++)
                for(int k = j + 1; k < nums.capacity(); k++)
                    if(nums[i] + nums[j] + nums[k] == 0){
                        vector<int> answer;
                        answer.push_back(nums[i]);
                        answer.push_back(nums[j]);
                        answer.push_back(nums[k]);
                        answers.push_back(answer);
                    }   
        
 
return answers;
    }
};

Submission Result: Wrong Answer More Details

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第二遍先對nums排序，代碼也是錯的：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> answers;
        std::sort(nums.begin(), nums.end());  //對nums排序
        int n=0,flag=0;
        if(nums.empty() || nums.capacity() < 3) return answers;
        for(int i=0; i < nums.capacity() - 2; i++){
            if(i>0 && nums[i] == nums[i - 1]) continue;   //如果當前元素等於前一個元素，跳過，防止出現重復配對
            for(int j = i + 1; j < nums.capacity() - 1; j++)
                for(int k = j + 1; k < nums.capacity(); k++)
                    if(nums[i] + nums[j] + nums[k] == 0){
                        vector<int> answer;
                        answer.push_back(nums[i]);
                        answer.push_back(nums[j]);
                        answer.push_back(nums[k]);
                        answers.push_back(answer);
                    }   
        }
        return answers;
    }
};

結果沒通過：

class Solution {
public:
   vector<vector<int> > threeSum(vector<int> &num) {
    
    vector<vector<int> > res;

    std::sort(num.begin(), num.end());

    for (int i = 0; i < num.size(); i++) {
        
        int target = -num[i];
        int front = i + 1;
        int back = num.size() - 1;

        while (front < back) {

            int sum = num[front] + num[back];
            
            // Finding answer which start from number num[i]
            if (sum < target)
                front++;

            else if (sum > target)
                back--;

            else {
                vector<int> triplet(3, 0);
                triplet[0] = num[i];
                triplet[1] = num[front];
                triplet[2] = num[back];
                res.push_back(triplet);
                
                // Processing duplicates of Number 2
                // Rolling the front pointer to the next different number forwards
                while (front < back && num[front] == triplet[1]) front++;

                // Processing duplicates of Number 3
                // Rolling the back pointer to the next different number backwards
                while (front < back && num[back] == triplet[2]) back--;
            }
            
        }

        // Processing duplicates of Number 1
        while (i + 1 < num.size() && num[i + 1] == num[i]) 
            i++;

    }
    
    return res;
    
}
};

15、3Sum