【LeetCode】#18四数之和(4Sum)
题目描述
给定一个包含 n 个整数的数组 nums 和一个目标值 target,判断 nums 中是否存在四个元素 a,b,c 和 d ,使得 a + b + c + d 的值与 target 相等?找出所有满足条件且不重复的四元组。
注意:
答案中不可以包含重复的四元组。
示例
给定数组 nums = [1, 0, -1, 0, -2, 2],和 target = 0。
满足要求的四元组集合为:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]
Description
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]
]
解法
class Solution{
public List<List<Integer>> fourSum(int[] nums, int target){
List<List<Integer>> res = new ArrayList<>();
Arrays.sort(nums);
for(int i=0; i<nums.length-3; i++){
for(int j=i+1; j<nums.length-2; j++){
if(j>(i+1) && nums[j]==nums[j+1]){
continue;
}
int left = j + 1;
int right = nums.length - 1;
while(left<right){
int sum = nums[i] + nums[j] + nums[left] + nums[right];
if(sum==target){
List<Integer> list = new ArrayList<>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[left]);
list.add(nums[right]);
if(!res.contains(list)){
res.add(list);
}
left++;
right--;
}else if(sum<target){
left++;
}else{
right--;
}
}
}
}
return res;
}
}