18. 4Sum（求数组中和为指定值的4个数）

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.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]
]

题目大意：给定一个整型数组和一个整数target，找到4个元素，和为target，返回这4个元素，要求返回的结果集不包含重复的方案。
解题思路：参照《15. 3Sum（求数组中和为0的3个数）》，先对数组排序，不同的是按排序结果每次锁定两个数nums[i]和nums[j]，然后用双指针思想从这两个数的右侧找到两个数，这两个数的和为target-nums[i]-nums[j]。
解题代码：（67ms，beats 61.09%）

class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> res = new ArrayList<>();
Arrays.sort(nums);
int len = nums.length;
for (int i = 0; i < len - 3; i++) {
for (int j = i + 1; j < len - 2; j++) {
int left = j + 1;
int right = len - 1;
int diff = target - nums[i] - nums[j];
while (left < right) {
if (nums[left] + nums[right] == diff) {
res.add(new ArrayList<>(Arrays.asList(nums[i], nums[j], nums[left], nums[right])));
left++;
while (left < len && nums[left] == nums[left - 1])
left++;
right--;
while (right > left && nums[right] == nums[right + 1])
right--;
} else if (nums[left] + nums[right] < diff) {
left++;
} else {
right--;
}
}
while (j < len - 3 && nums[j] == nums[j + 1]) {
j++;
}
}
while (i < len - 4 && nums[i] == nums[i + 1]) {
i++;
}
}
return res;
}
}