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代码随想录第二十天

哈希表-四数之和

题目链接

这道题采用双指针的方法,首先顺序依次是i,j,left,right。和三数之和是一样的操作

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class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> result;
sort(nums.begin(), nums.end());
for(int i = 0; i < nums.size(); i++){

if(i > 0 && nums[i] == nums[i - 1])continue;

for(int j = i + 1; j < nums.size(); j++){

if(j > i + 1 && nums[j] == nums[j - 1])continue;

int left = j + 1;
int right = nums.size() - 1;

while(left < right){
// if(nums[i] + nums[j] + nums[left] + nums[right] > target)right--;//会溢出
if(nums[i] + nums[j] > target - (nums[left] + nums[right]))
right--;


else if(nums[i] + nums[j] < target - (nums[left] + nums[right]))
left++;

else{
result.push_back(vector<int>{nums[i], nums[j], nums[left], nums[right]});
while(left < right && nums[right] == nums[right - 1]) right--;
while(left < right && nums[left] == nums[left + 1]) left++;

right--;
left++;
}


}
}

}
return result;
}
};
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def fourSum(self, nums: List[int], target: int) -> List[List[int]]:

nums.sort()
n = len(nums)
res = []
for i in range(n):
if i > 0 and nums[i] == nums[i - 1]: continue
for k in range(i+1, n):
if k > i + 1 and nums[k] == nums[k-1]: continue
p = k + 1
q = n - 1

while p < q:
if nums[i] + nums[k] + nums[p] + nums[q] > target: q -= 1
elif nums[i] + nums[k] + nums[p] + nums[q] < target: p += 1
else:
res.append([nums[i], nums[k], nums[p], nums[q]])
while p < q and nums[p] == nums[p + 1]: p += 1
while p < q and nums[q] == nums[q - 1]: q -= 1
p += 1
q -= 1
return res