18. 4Sum

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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: The solution set must not contain duplicate quadruplets.

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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]
]

解法1： O(N^3)

和3Sum一样的思路，最外层遍历第一个数，第二层遍历第二个数，之后用双指针像中间扫描。遇到加和为target的就推入list中。要注意的是需要去重。那么每一次去重的时候只需要跳过和前一个一样的数就可以了。
C++

1

Java

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public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        int n = nums.length;
        if (n < 4) {
            return res;
        }
        Arrays.sort(nums);      // need to sort first
        for (int i = 0; i < n - 3; i++) {
            if (i != 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            for (int j = i + 1; j < n - 2; j++) {
                if (j != i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
                int k = j + 1;
                int m = n - 1;
                while (k < m) {
                    while (k != j + 1 && k < m && nums[k] == nums[k - 1]) {
                        k++;    // remove duplicates
                    }
                    while (m != n - 1 && k < m && nums[m] == nums[m + 1]) {
                        m--;
                    }
                    if (k < m) {
                        int sum = nums[i] + nums[j] + nums[k] + nums[m];
                        if (sum == target) {
                            List<Integer> temp = new ArrayList<Integer>();
                            temp.add(nums[i]);
                            temp.add(nums[j]);
                            temp.add(nums[k]);
                            temp.add(nums[m]);
                            res.add(temp);
                            k++;
                            m--;
                        } else if (sum < target) {
                            k++;
                        } else {
                            m--;
                        }
                    }
                }
            }
        }
        return res;
    }
}