给你一个由 n
个整数组成的数组 nums
,和一个目标值 target
。请你找出并返回满足下述全部条件且不重复的四元组 [nums[a], nums[b], nums[c], nums[d]]
(若两个四元组元素一一对应,则认为两个四元组重复):
0 <= a, b, c, d < n
a
、b
、c
和d
互不相同nums[a] + nums[b] + nums[c] + nums[d] == target
你可以按 任意顺序 返回答案 。
示例 1:
输入:nums = [1,0,-1,0,-2,2], target = 0 输出:[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> results = new ArrayList<>();
Arrays.sort(nums);
int numSize = nums.length;
int left, right;
// the first element circle
for(int a=0; a<numSize-3; a++) {
// skip the same tuple if existed
if (a != 0 && nums[a] == nums[a-1]) continue;
// the second element circle
for(int b=a+1; b<numSize-2; b++) {
// skip the same element
if(b!= a+1 && nums[b] == nums[b-1]) continue;
// the third element circle
left = b+1;
right = numSize -1;
while(left < right) {
long sum = (long)nums[a] + nums[b] + nums[left] +nums[right];
if (sum == target) {
results.add(Arrays.asList(nums[a], nums[b], nums[left], nums[right]));
// skip the same element from left side
while(left < right && nums[left] == nums[left+1]) left++;
// skip the same element from right side
while(left < right &&nums[right-1] == nums[right]) right--;
left++;
right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
}
return results;
}
}
解法参考: