### 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)

``````import java.util.*;

/**
* Created by gzdaijie on 16/5/18
* 类似于15题——3sum, 多加一层循环
* 先以O(N*N)的复杂度枚举可能出现的2个数的组合, 再按照3sum的解法寻找和为target的另外2个数
* 总的时间复杂度O(N*N*N)
*/
public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
int len = nums.length;
List<List<Integer>> result = new ArrayList<List<Integer>>();
if (len < 4) return result;

Arrays.sort(nums);
for (int i = 0; i < len - 3; i++) {
if (i != 0) {
while (i < len - 3 && nums[i] == nums[i - 1]) i++;
}

for (int j = i + 1; j < len - 2; j++) {
if (j != i + 1) {
while (j < len - 2 && nums[j] == nums[j - 1]) j++;
}

int left = j + 1;
int right = nums.length - 1;
int sum = nums[i] + nums[j];
while (left < right) {
int total = sum + nums[left] + nums[right];
if (total == target) {
++left;
while (left < right && nums[left] == nums[left - 1]) ++left;
} else if (total < target) {
++left;
} else {
--right;
}

}
}
}
return result;
}
}
``````

updated 2016-05-18 18:52:49