LeetCode--018--四数之和(java)

时间:2022-04-10 09:30:56

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]
]
和三数之和的区别就是,在外层多了一层for循环。
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);
for(int i = 0;i < nums.length - 3;i++){
if(i > 0 && nums[i] == nums[i-1])continue;
for(int j = i + 1;j < nums.length - 2;j++){
if(j > i + 1 && nums[j] == nums[j-1]) continue;
int low = j + 1,high = nums.length - 1;
while(low < high){
int sum = nums[i] + nums[j] + nums[low] + nums[high];
if(sum == target){
res.add(Arrays.asList(nums[i],nums[j],nums[low],nums[high]));
while(low < high && nums[low] == nums[low+1])low++;
while(low < high && nums[high] == nums[high-1]) high--;
low++;
high--;
}else if(sum < target){
low++;
}else high--;
}
}
}
return res;
}
}

2019-04-17 21:23:25

用python按上面的写一遍,发现根本过不了,要提前停止搜索,所以又加了四个if

 class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
n = len(nums)
if n<4:
return []
nums.sort()
res = []
for i in range(n-3):
if i>0 and nums[i]==nums[i-1]:continue
if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target):
break
if(nums[i]+nums[-1]+nums[-2]+nums[-3]<target):
continue
for j in range(i+1,n-2):
if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target):
break
if(nums[i]+nums[j]+nums[-1]+nums[-2]<target):
continue
if j>i+1 and nums[j]==nums[j-1]:continue
L = j+1
R = n-1
while L<R:
print(R)
temp = nums[i]+nums[j]+nums[L]+nums[R]
if temp == target:
res.append([nums[i],nums[j],nums[L],nums[R]])
while L<R and nums[L]==nums[L+1]:
L+=1
while L<R and nums[R]==nums[R-1]:
R-=1
L+=1
R-=1
elif temp < target:
L+=1
else:
R-=1
return res

2019-11-30 08:32:54