454.四数相加II
关键点：减少复杂度判断 a+b 是否 等于 -(d+e).求和类的题目，利用好相反数。
class Solution:
def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
record = {}
for i in nums1:
for j in nums2:
key = i + j
record[key] = record.get(key, 0) + 1
count = 0
for k in nums3:
for l in nums4:
target = -(k+l)
if target in record:
count += record.get(target)
return count
383. 赎金信
1、字典
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
record = {}
for i in magazine:
record[i] = record.get(i, 0) + 1
for j in ransomNote:
if j not in record or record.get(j) == 0:
return False
else:
record[j] = record.get(j) - 1
return True
2、数组
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
record = [0]*26
for i in ransomNote:
record[ord(i)-ord("a")] += 1
for j in magazine:
record[ord(j)-ord("a")] -= 1
for k in record:
if k > 0:
return False
return True
15.三数之和
注意点：对于双指针需要添加到结果集后进行一个去重
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
res = []
n = len(nums)
nums.sort()
for i in range(n):
if nums[i] > 0:
return res
if i > 0 and nums[i] == nums[i-1]: # 需要加入到结果集后再对 nums[i] 去重
continue
left, right = i+1, n-1
while left < right:
sum_val = nums[i] + nums[left] + nums[right]
if sum_val < 0:
left += 1
elif sum_val > 0:
right -= 1
else:
res.append([nums[i], nums[left], nums[right]])
while left < right and nums[left] == nums[left+1]: # 对 nums[left] 去重
left += 1
while left < right and nums[right] == nums[right-1]: # 对 nums[right] 去重
right -= 1
left += 1
right -= 1
return res
18. 四数之和
注意点：剪枝的位置。第二层的不能return res
class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
res = []
n = len(nums)
nums.sort()
for k in range(n):
if nums[k] > target and target > 0: # 剪枝
break
if k > 0 and nums[k] == nums[k-1]: #去重
continue
for i in range(k+1, n):
if nums[i] + nums[k] > target and target > 0: # 剪枝
break
if i > k+1 and nums[i] == nums[i-1]: #去重
continue
left, right = i+1, n-1
while left < right:
sum_val = nums[k] + nums[i] + nums[left] + nums[right]
if sum_val < target:
left += 1
elif sum_val > target:
right -= 1
else:
res.append([nums[k], nums[i], nums[left], nums[right]])
while left < right and nums[left] == nums[left+1]: # 去重
left += 1
while left < right and nums[right] == nums[right-1]: # 去重
right -= 1
left += 1
right -= 1
return res