这一部分开始,我们应用双指针及哈希等常见的简单的算法,解决一些字符串的难题。
345. Reverse Vowels of a String
Write a function that takes a string as input and reverse only the vowels of a string.
Example 1:
Input: "hello" Output: "holle"
Example 2:
Input: "leetcode" Output: "leotcede"
Note:
The vowels does not include the letter "y".
题目解析:
双指针在字符串的应用也很简单,这道题难度也是easy。需要说明的是字符串中字符的 交换只能用切片,下面的代码则是先将字符串转为列表后再操作的。
class Solution:
def reverseVowels(self, s):
"""
:type s: str
:rtype: str
"""
i = 0
j = len(s) - 1
l = list(s)
while i < j:
while not l[i] in "aeiouAEIOU" and i < j:
i += 1
while not l[j] in "aeiouAEIOU" and i < j:
j -= 1
if i >= j:
break
l[i], l[j] = l[j], l[i]
# s = s[:i] + s[j] + s[i+1:j] + s[i] + s[j+1:] # 切片的话这样写
i += 1
j -= 1
return ''.join(l)
3. Longest Substring Without Repeating Characters(最长无重复字符子串)
Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3.
题目解析:
与上一题想比,两个指针都是从头扫描,在滑动窗内遇到重复的字符,将前面的舍弃掉,并重置n,此时不可能使n增长,所以无需判断;无重复的则添加增长字符串,并判断是否最长。
class Solution:
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
longest = []
length_max = 0
n = 0
for x in s:
if x in longest:
ix = longest.index(x)
longest = longest[(ix+1):] # 舍弃前面的
longest.append(x)
n = len(longest)
# print(n,longest)
else:
longest.append(x)
n += 1
# print(n, longest)
if n > length_max:
length_max = n
return length_max