leetcode #35 搜索插入位置
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
if len(nums) == 0:
return 0
left = 0
right = len(nums) - 1
if target > nums[-1]:
return len(nums)
if target < nums[0]:
return 0
while left <= right:
mid = (left + right)//2
if nums[mid] == target:
return mid
elif nums[mid] > target:
right = mid - 1
else:
left = mid + 1
return left
leetcode #202快樂數(shù)
class Solution:
def isHappy(self, n: int) -> bool:
def get_next(n):
total_sum = 0
while n > 0:
n, digit = divmod(n, 10)
total_sum += digit ** 2
return total_sum
seen = set()
while n != 1 and n not in seen:
seen.add(n)
n = get_next(n)
return n == 1
leetcode #205 同構(gòu)字符串
class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
if len(set(s)) != len(set(t)):
return False
dic_s = dict()
lst_s = []
dic_t = dict()
lst_t = []
new = 1
for char in s:
if char not in dic_s:
dic_s.setdefault(char,new)
lst_s.append(new)
new += 1
else:
lst_s.append(dic_s[char])
new = 1
for char in t:
if char not in dic_t:
dic_t.setdefault(char,new)
lst_t.append(new)
new += 1
else:
lst_t.append(dic_t[char])
if lst_s == lst_t:
return True
else:
return False
leetcode #242 有效的字母異位詞
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
dic_s = dict()
dic_t = dict()
for char in s:
dic_s.setdefault(char,0)
dic_s[char] += 1
for char in t:
dic_t.setdefault(char,0)
dic_t[char] += 1
if dic_s == dic_t:
return True
else:
return False
leetcode #290 單詞規(guī)律
class Solution:
def wordPattern(self, pattern: str, str: str) -> bool:
s_lst = str.split(' ')
dic_p = dict()
dic_s = dict()
lst_s = []
lst_p = []
new = 1
for char in pattern:
if char not in dic_p:
dic_p.setdefault(char, new)
lst_p.append(new)
new += 1
else:
lst_p.append(dic_p[char])
new = 1
for char in s_lst:
if char not in dic_s:
dic_s.setdefault(char, new)
lst_s.append(new)
new += 1
else:
lst_s.append(dic_s[char])
if lst_p == lst_s:
return True
else:
return False
leetcode #349 兩個數(shù)組的交集
class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
lst = []
nums1 = set(nums1)
nums2 = set(nums2)
for val in nums1:
if val in nums2:
lst.append(val)
return lst
leetcode #350 兩個數(shù)組的交集2
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
if len(nums1) > len(nums2):
return self.intersect(nums2, nums1)
m = collections.Counter()
for num in nums1:
m[num] += 1
intersection = list()
for num in nums2:
if (count := m.get(num, 0)) > 0:
intersection.append(num)
m[num] -= 1
if m[num] == 0:
m.pop(num)
return intersection
leetcode #410 分割數(shù)組的最大值
class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
def check(x: int) -> bool:
total, cnt = 0, 1
for num in nums:
if total + num > x:
cnt += 1
total = num
else:
total += num
return cnt <= m
left = max(nums)
right = sum(nums)
while left < right:
mid = (left + right) // 2
if check(mid):
right = mid
else:
left = mid + 1
return left
leetcode #451 根據(jù)字符出現(xiàn)次數(shù)排序
class Solution:
def frequencySort(self, s: str) -> str:
set_s = set(s)
lst = []
for char in set_s:
count = 0
for i in s:
if char == i:
count += 1
lst.append((count, char))
lst = sorted(lst, key = lambda x:x[0], reverse = True)
res = ''
for count, char in lst:
res += char*count
return res
leetcode #540 有序數(shù)組中的單一元素
class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
def rec(left, right):
if left == right:
return nums[left]
mid = (left + right)//2
if nums[mid] != nums[mid-1] and nums[mid] != nums[mid+1]:
return nums[mid]
if nums[mid] == nums[mid - 1]:
if (mid-1-left) % 2 != 0:
return rec(left, mid-2)
else:
return rec(mid+1, right)
else:
if (right-mid-1) % 2 != 0:
return rec(mid+2, right)
else:
return rec(left, mid-1)
return rec(0, len(nums)-1)
