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)

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class Solution {
public:
    /*
     * 固定两个点,然后使用two sum 求解
     * 注意去掉重复的点
     */
    void two_sum(vector<int>& nums, vector<vector<int>>& res, int start_index, int last_index, int target) {
        int i = start_index + 1;
        int j = last_index - 1;
        while (i < j) {
            if (nums[i] + nums[j] == target) {
                vector<int> tmp;
                tmp.push_back(nums[start_index]);
                tmp.push_back(nums[i]);
                tmp.push_back(nums[j]);
                tmp.push_back(nums[last_index]);
                res.push_back(tmp);
                cout << nums[start_index] << nums[i] << nums[j] << nums[last_index] << endl;
                i++;
                while (nums[i] == nums[i-1]) {
                    i++;
                }
                j--;
                while (nums[j] == nums[j+1]) {
                    j--;
                }
            } else if (nums[i] + nums[j] > target) {
                j--;
            } else {
                i++;
            }
        }
    }

    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int size = nums.size();
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        int i;
        int j;
        for (i=0; i<size-2; i++) {
            //过滤掉相同的左边点,防止出现重复的结果
            if (i > 0 && nums[i] == nums[i-1]) {
                continue;
            }
            for (j=size-1; j>i+1; j--) {
                if (j < size-1 && nums[j] == nums[j+1]) {
                    continue;
                }
                two_sum(nums, res, i, j, target- nums[i] - nums[j]);
            }
        }
        return res;
    }
};