[LeetCode] 001: 3Sum

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:

• 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.

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

[Analysis]
对数组进行排序，利用3个指针，2个在数组前端，1个在后端，利用两边夹逼的方法进行搜索。
注意：在移动指针的时候，过滤相同的数值。

[Solution]

class Solution {
   
public:
    vector<vector<int> > threeSum(vector<int> &num) {
   
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        // result
        vector<vector<int> > result;
        
        // sort the array
        sort(num.begin(), num.end());
        
        // search
        int i = 0, j, k;
        while(i < num.size()){
   
		    // ignore same num[i]
		    while(i > 0 && num[i] == num[i-1]){
   
			    i++;
		    }

		    // initial j and k
		    j = i + 1;
		    k = num.size() - 1;
		
		    // search
	    	while(j < k){
   
		    	// ignore same num[k]
		    	while(k < num.size()-1 && num[k] == num[k+1] && k > j){
   
		    		k--;
		    	}

		    	// ignore same num[j]
		    	while(j > i+1 && num[j] == num[j-1] && j < k){
   
		    		j++;
		    	}
		    	if(j >= k)break;
		    	
		    	// equals to 0
		    	if(num[i] + num[j] + num[k] == 0){
   
		    	    vector<int> tmp;
		    	    tmp.push_back(num[i]);
		    	    tmp.push_back(num[j]);
		    	    tmp.push_back(num[k]);
		    	    
		    	    // insert into result
		    	    result.push_back(tmp);
		    	    j++;
		    	}
		    	else if(num[i] + num[j] + num[k] > 0){
   
		    	    k--;
		    	}
		    	else{
   
		    	    j++;
		    	}
		    }
		    i++;
        }
        return result;
    }
};