标签：leetcode ret script turn solution none ble arch problem
class Solution:
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
ss = list(s)
tt = list(t)
ss.sort()
tt.sort()
return ss == tt
"""
输入："anagram"、"nagaram"
输出：true
Runtime: 32 ms
"""
排序方法简写如下：
class Solution:
def isAnagram(self, s, t):
return sorted(list(s)) == sorted(list(t))
class Solution:
def isAnagram(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
dict1 = {} # 用字典来维护字符的数量
dict2 = {}
for ch in s:
dict1[ch] = dict1.get(ch, 0) + 1 # 没有就新建，有就加1
for ch in t:
dict2[ch] = dict2.get(ch, 0) + 1
return dict1 == dict2
"""
输入："anagram"，"nagaram"
输出：true
Runtime: 32 ms
"""
列表有下列特性：
（1）解法一：线性查找查找，O(mn)
class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
for line in matrix:
if target in line:
return True
return False
（2）解法二：二分查找O(logn)
class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
h = len(matrix) # 高
if h == 0:
return False
w = len(matrix[0]) # 列
if w == 0:
return False
left = 0
right = w * h - 1
while left <= right:
mid = ((left + right)) // 2
i = mid // w
j = mid % w
if matrix[i][j] == target:
return True
elif matrix[i][j] > target:
right = mid - 1
else:
left = mid + 1
else:
return False
保证肯定仅有一个结果。
leetcode地址：https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/
例如，列表[1,2,5,4]与目标整数3，1+2=3，结果为(0,1).
（1）方法一：通过二分查找，找到需要的数字。时间复杂度：O(nlogn)
首先确定第一个数，再通过给定整数确定要查找的数，通过二分查找到需要的数。
class Solution:
def binary_search(self, li, left, right,val):
"""
二分查找
:param li: 输入的列表
:param val: 输入的待查找的值
:return:
"""
while left <= right: # 说明候选区有值
mid = (left + right) // 2 # 因为是下标， 因此要整除2
if li[mid] == val:
# 找到待查找的值返回index
return mid
elif li[mid] > val:
# 待查找的值在mid左侧
right = mid - 1 # 更新候选区
else: # li[mid] < val
# 待查找的值在mid右侧
left = mid + 1 # 更新候选区
else:
# 没有找到
return None
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
for i in range(len(nums)):
a = nums[i]
b = target - a
if b >= a:
j = self.binary_search(nums, i+1, len(nums)-1, b)
else:
j = self.binary_search(nums, 0, i-1, b)
if j:
break
return sorted([i+1,j+1])
（2）方法二：针对已经排好序的列表
leetcode地址：https://leetcode.com/problems/two-sum/description/
class Solution:
def binary_search(self, li, left, right,val):
"""
二分查找
:param li: 输入的列表
:param val: 输入的待查找的值
:return:
"""
while left <= right: # 说明候选区有值
mid = (left + right) // 2 # 因为是下标， 因此要整除2
if li[mid][0] == val:
# 找到待查找的值返回index
return mid
elif li[mid][0] > val:
# 待查找的值在mid左侧
right = mid - 1 # 更新候选区
else: # li[mid] < val
# 待查找的值在mid右侧
left = mid + 1 # 更新候选区
else:
# 没有找到
return None
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
new_nums = [[num, i] for i,num in enumerate(nums)]
new_nums.sort(key=lambda x:x[0])
for i in range(len(new_nums)):
a = new_nums[i][0] # 数
b = target - a
if b >= a:
j = self.binary_search(new_nums, i+1, len(new_nums)-1, b)
else:
j = self.binary_search(new_nums, 0, i-1, b)
if j:
break
return sorted([new_nums[i][1], new_nums[j][1]])
标签：leetcode ret script turn solution none ble arch problem
原文地址：https://www.cnblogs.com/xiugeng/p/9745743.html
