454. 四數(shù)相加 II
- 思路
- example
- 四個數(shù)組相同長度
- 第一步：brute-force:
- hash: dict
- record[num] = freq
第二步：
第三步：(nums1[i]+nums2[j]) + (nums3[k]+nums4[l])，
- 復(fù)雜度. 時間：
, 空間:
class Solution:
def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
n = len(nums1)
record = collections.defaultdict(int)
for i in range(n):
for j in range(n):
record[nums1[i]+nums2[j]] += 1
res = 0
for k in range(n):
for l in range(n):
if -(nums3[k] + nums4[l]) in record:
res += record[-(nums3[k] + nums4[l])]
return res
class Solution:
def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
record = collections.defaultdict(int)
for num1 in nums1:
for num2 in nums2:
record[num1+num2] += 1
res = 0
for num3 in nums3:
for num4 in nums4:
if record[-(num3+num4)] > 0:
res += record[-(num3+num4)] # !!!
return res
383. 贖金信
- 思路
- example
- hash: dict
- record[ch] = freq
先遍歷模式串magazine
- 復(fù)雜度. 時間：O(n), 空間: O(n)
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
table = collections.defaultdict(int)
for ch in magazine:
table[ch] += 1
for ch in ransomNote:
table[ch] -= 1
if table[ch] < 0:
return False
return True
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
record = collections.defaultdict(int)
for ch in magazine:
record[ch] += 1
for ch in ransomNote:
record[ch] -= 1
if record[ch] < 0:
return False
return True
15. 三數(shù)之和
- 思路
- example
- a + b + c = 0
- 用hash的話去重問題更麻煩。
- 雙指針, --><--
-
去重：(用排序配合)
- a: nums[i] == nums[i-1]
- b and c: while-loop moving left and right
-
去重：(用排序配合)
- 復(fù)雜度. 時間：
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
n = len(nums)
res = []
for i in range(n):
if i > 0 and nums[i] == nums[i-1]: # 去重１
continue
target = -nums[i]
left, right = i + 1, n - 1
while left < right:
sum_ = nums[left] + nums[right]
if sum_ == target:
res.append([nums[i], nums[left], nums[right]])
left += 1
while left < right and nums[left] == nums[left-1]:
left += 1
right -= 1
while left < right and nums[right] == nums[right+1]:
right -= 1
elif sum_ < target:
left += 1
else:
right -= 1
return res
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
n = len(nums)
nums.sort()
res = []
for i in range(n):
if i > 0 and nums[i] == nums[i-1]:
continue
target = -nums[i]
left, right = i+1, n-1
while left < right:
sum_ = nums[left] + nums[right]
if sum_ < target:
left += 1
elif sum_ > target:
right -= 1
else:
res.append([nums[i], nums[left], nums[right]])
left += 1
while left < right and nums[left] == nums[left-1]:
left += 1
right -= 1
while left < right and nums[right] == nums[right+1]:
right -= 1
return res
18. 四數(shù)之和
- 思路
- example
- 雙指針，三數(shù)之和推廣
- 去重，三個層面: a, b, 以及c and d
- 復(fù)雜度. 時間：
, 空間:
class Solution:
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]: # 同層去重
continue
for j in range(i+1, n):
if j > i + 1 and nums[j] == nums[j-1]: # 同層去重
continue
target2 = target - (nums[i] + nums[j])
left, right = j + 1, n - 1
while left < right:
sum_ = nums[left] + nums[right]
if sum_ == target2:
res.append([nums[i], nums[j], nums[left], nums[right]])
left += 1
while left < right and nums[left] == nums[left-1]:
left += 1
right -= 1
while left < right and nums[right] == nums[right+1]:
right -= 1
elif sum_ < target2:
left += 1
else:
right -= 1
return res
class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
n = len(nums)
res = []
nums.sort()
for i in range(n):
if i > 0 and nums[i] == nums[i-1]:
continue
for j in range(i+1, n):
if j > i+1 and nums[j] == nums[j-1]:
continue
target2 = target-(nums[i] + nums[j]) # !!!
left, right = j+1, n-1
while left < right:
sum_ = nums[left] + nums[right]
if sum_ < target2:
left += 1
elif sum_ > target2:
right -= 1
else:
res.append([nums[i], nums[j], nums[left], nums[right]])
left += 1
while left < right and nums[left] == nums[left-1]:
left += 1
right -= 1
while left < right and nums[right] == nums[right+1]:
right -= 1
return res
