[LeetCode] 003: 4Sum

[Problem]

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

[Analysis]
与3Sum类似，这里使用4个指针，前2个不动。后2个从两边进行夹逼。

[Solution]

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // 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, l;
        while(i < num.size()){
		    // ignore same num[i]
		    while(i > 0 && num[i] == num[i-1]){
			    i++;
		    }

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

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

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

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