题目描述:
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] ]
中文理解:
给定一个数组和target,找出数组中左右和为target的四个数字组合,返回这些所有组合。
解题思路:
首先将数组排序,然后外层使用i,j双层循环,j>i,内层使用两个指针start,end,将题目转换为3sum来解决。
代码(java):
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> res=new ArrayList();
if(nums.length<4)return res;
Arrays.sort(nums);
int flag=0;
for(int i=0;i<nums.length-3;i++){
if(i>0&&nums[i]==nums[i-1])continue;
if(nums[i]==-1)System.out.println(nums[i]);
for(int j=i+1;j<nums.length-2;j++){
if(j>i+1 && nums[j]==nums[j-1])continue;
if(nums[i]==-1)System.out.println(nums[i]+" "+nums[j]);
int start=j+1,end=nums.length-1;
while(start<end){
if(nums[i]==-1)System.out.println(nums[i]+" "+nums[j]+" "+nums[start]+" "+nums[end]);
int temp=nums[i]+nums[j]+nums[start]+nums[end];
if(temp<target){
start++;
}
else if(temp>target){
end--;
}
else if(temp==target){
res.add(Arrays.asList(nums[i],nums[j],nums[start],nums[end]));
while(start<end&&nums[start]==nums[start+1])start++;
while(start<end && nums[end]==nums[end-1])end--;
start++;
end--;
}
}
}
}
return res;
}
}