leetcode: 4Sum

http://oj.leetcode.com/problems/4sum/

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)

思路：

把3Sum的代码扩展一下搞定。

class Solution {
public:
    void threeSum(vector<int> &num, int start_index, int target, int v1, vector<vector<int> > &result) {
        int size = num.size();
        
        for (int i = start_index; i < size - 2; ) {
            int j = i + 1, k = size - 1, tmp;
            
            while (j < k) {
                int sum = v1 + num[i] + num[j] + num[k];
                
                if (target == sum) {
                    vector<int> r;
                    
                    r.push_back(v1);
                    r.push_back(num[i]);
                    r.push_back(num[j]);
                    r.push_back(num[k]);
                    result.push_back(r);
                }
                
                if (sum < target) {
                    tmp = j + 1;
                    
                    while ((tmp < k) && (num[j] == num[tmp])) {
                        ++tmp;
                    }
                    
                    j = tmp;
                }
                else {
                    tmp = k - 1;
                    
                    while ((tmp > j) && (num[tmp] == num[k])) {
                        --tmp;
                    }
                    
                    k = tmp;
                }
            }
            
            tmp = i + 1;
            
            while ((tmp < (size - 2)) && (num[i] == num[tmp])) {
                ++tmp;
            }
            
            i = tmp;
        }
    }
    
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        vector<vector<int> > result;
        int size = num.size(), tmp;
        
        sort(num.begin(), num.end());
        
        for (int i = 0; i < (size - 3); ) {
            threeSum(num, i + 1, target, num[i], result);
            
            tmp = i + 1;
            
            while ((tmp < (size - 3)) && (num[i] == num[tmp])) {
                ++tmp;
            }
            
            i = tmp;
        }
        
        return result;
    }
};