leetcode-18. 4Sum

https://leetcode.com/problems/4sum/#/description

问题描述：

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]

]

思路解析：

参考这两道题：

http://blog.csdn.net/u013275928/article/details/71440744

http://blog.csdn.net/u013275928/article/details/71698864

设立四个指针，注意边界由于是四个值，集合中可能有重复的元组出现，所以需要list的contains()来判断当前符合条件的元组是否在集合中出现过，若未出现则加进来。然后移动指针。

代码如下：

public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {

List<List<Integer>> result=new ArrayList<>();

if(nums==null || nums.length<4)
{
return result;
}

Arrays.sort(nums);

for(int i=0;i<nums.length-3;i++ )
{
for(int j=i+1;j<nums.length-2;j++)
{
int h=j+1;
int k=nums.length-1;
while(h<k)
{
int sum=nums[i]+nums[j]+nums[h]+nums[k];

if(sum==target) {

// List<Integer> temp=new ArrayList<Integer>();
// temp.add();
if(!result.contains(Arrays.asList(nums[i],nums[j],nums[h],nums[k])))
{
result.add(Arrays.asList(nums[i],nums[j],nums[h],nums[k]));
}

h++;
k--;

}
else if(sum>target)k--;
else h++;

}

}

{

}

}

return result;
}
}