判定是否互为字符重排
给定两个字符串 s1 和 s2,请编写一个程序,确定其中一个字符串的字符重新排列后,能否变成另一个字符串。
示例 1:
- 输入: s1 = "abc", s2 = "bca"
- 输出: true
示例 2:
- 输入: s1 = "abc", s2 = "bad"
- 输出: false
示例代码1:
# 位运算
class Solution(object):
def CheckPermutation(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: bool
"""
if len(s1) != len(s2):
return False
res = 0
for i in range(len(s1)):
res += 1 << ord(s1[i])
res -= 1 << ord(s2[i])
return res == 0
a = Solution()
# b = a.CheckPermutation('abc', 'bca')
b = a.CheckPermutation('abd', 'bca')
print(b)
示例代码2:
# 字符串排序后比较
class Solution(object):
def CheckPermutation(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: bool
"""
return sorted(s1) == sorted(s2)
a = Solution()
b = a.CheckPermutation('abc', 'bca')
# b = a.CheckPermutation('abd', 'bca')
print(b)
示例代码3:
# 利用集合元素的不可重复性
class Solution(object):
def CheckPermutation(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: bool
"""
return set(s1) == set(s2)
a = Solution()
# b = a.CheckPermutation('abc', 'bca')
b = a.CheckPermutation('abd', 'bca')
print(b)
运行效果:
解题思路:
详细了解python 中的ord()函数和chr()函数,查看博文:https://blog.csdn.net/weixin_44799217/article/details/112333924
详细了解Python 位运算符,查看博文:https://blog.csdn.net/weixin_44799217/article/details/111715689
示例代码1:
- 思路:位运算
- 时间复杂度: O(N)
- 空间复杂度: O(1)
示例代码3:
- 将字符串转为集合
- 利用集合元素的不可重复性