18. 4Sum

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums 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.

Example:
Given array nums = [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]
]

难度：medium

题目：
给定一整数数组和一目标整数，是否存在4个数的和为指定整数？ 找出所有这样不同的4元组。
注意：
答案一定不能包含重复的四元组。

思路：
数组排序，固定一个数，然后降级为3sum

Runtime: 34 ms, faster than 78.86% of Java online submissions for 4Sum.
Memory Usage: 31.7 MB, less than 13.82% of Java online submissions for 4Sum.

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList();
        Arrays.sort(nums);
        int numsLen = nums.length;
        for (int k = 0; k < numsLen; k++) {
            if (0 == k || nums[k - 1] != nums[k]) {
                for (int i = k + 1; i < numsLen; i++) {
                    if ((k + 1) == i || nums[i - 1] != nums[i]) {
                        for (int left = i + 1, right = numsLen - 1; left < right; left++, right--) {
                            int sum = nums[left] + nums[right] + nums[i] + nums[k];
                            if (target == sum 
                                && ((i + 1) == left || nums[left - 1] != nums[left])
                                && ((numsLen - 1) == right || nums[right] != nums[right + 1])
                               ) {
                                result.add(Arrays.asList(nums[k], nums[i], nums[left], nums[right]));
                            } else if (sum > target) {
                                left--;
                            } else {
                                right++;
                            }
                        }
                    }        
                }
            }
        }

        return result;
    }
}