LeetCode OJ：4Sum（4数字之和）

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)

和前面的思路基本上是相同的，这里同样容易遇见相同的数，这个时候需要跳过，代码如下：

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int sz = nums.size();
        vector<vector<int>> ret;
        for(int i = 0; i < sz; ++i){
            if(i != 0 && nums[i] == nums[i - 1])
                continue;
            for(int j = i + 1; j < sz; ++j){
                if(j > i + 1 && nums[j] == nums[j - 1])
                    continue;
                for(int beg = j + 1, end = sz - 1; beg < end;){
                    while(beg > j + 1 && nums[beg] == nums[beg - 1])
                        beg++;
                    while(end < sz - 1 && nums[end] == nums[end + 1])
                        end--;
                    if(beg >= end) break;
                    int sum = nums[i] + nums[j] + nums[beg] + nums[end];
                    if(sum == target){
                        vector<int> tmp;
                        tmp.push_back(nums[i]);
                        tmp.push_back(nums[j]);
                        tmp.push_back(nums[beg]);
                        tmp.push_back(nums[end]);
                        ret.push_back(tmp);
                        beg++, end--;
                    }else if(sum < target){
                        beg++;
                    }else{
                        end--;
                    }
                }
            }
        } 
        return ret;
    }
};