小知识，大挑战！本文正在参与“程序员必备小知识”创作活动
描述
Given a 0-indexed string word and a character ch, reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing.
- For example, if word = "abcdefd" and ch = "d", then you should reverse the segment that starts at 0 and ends at 3 (inclusive). The resulting string will be "dcbaefd".
Return the resulting string.
Example 1:
Input: word = "abcdefd", ch = "d"
Output: "dcbaefd"
Explanation: The first occurrence of "d" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd".
Example 2:
Input: word = "xyxzxe", ch = "z"
Output: "zxyxxe"
Explanation: The first and only occurrence of "z" is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe".
Example 3:
Input: word = "abcd", ch = "z"
Output: "abcd"
Explanation: "z" does not exist in word.
You should not do any reverse operation, the resulting string is "abcd".
Note:
1 <= word.length <= 250
word consists of lowercase English letters.
ch is a lowercase English letter.
解析
根据题意，就是给出了一个英文字符串 word ，又给出了一个字符 ch ，如果当 ch 存在于 word 中，那就让 word 从头开始到这个 ch 位置的子字符串进行反转，最后返回处理后的字符串结果，如果 ch 没有存在于 word 中，那么不做任何的操作。思路很简单：
- 如果 ch 不在 word 中，直接返回 word
- 否则找出 ch 在 word 中的索引 i ，直接返回 word[:i+1][::-1] + word[i+1:] 即可
解答
class Solution(object):
def reversePrefix(self, word, ch):
"""
:type word: str
:type ch: str
:rtype: str
"""
if ch not in word:
return word
i = word.index(ch)
return word[:i+1][::-1] + word[i+1:]
运行结果
Runtime: 29 ms, faster than 29.96% of Python online submissions for Reverse Prefix of Word.
Memory Usage: 13.4 MB, less than 62.55% of Python online submissions for Reverse Prefix of Word.
解析
其实上面的方法就是用到了内置函数，直接遍历也是可以的，直接遍历 word ，每次将字符添加到 result 的开头，一直等找到 ch 为止，然后将当前的 ch 也加到 result 的最前头，最后将剩下的部分追加到 result 结果后面就可以了。这个方法有点凑数的嫌疑，反正题很简单，代码随便浪就行了。
解答
class Solution(object):
def reversePrefix(self, word, ch):
"""
:type word: str
:type ch: str
:rtype: str
"""
if ch not in word:
return word
result = ''
i = 0
for c in word:
if c != ch:
result = c + result
i += 1
else:
break
result = word[i] + result
return result + word[i+1:]
运行结果
Runtime: 31 ms, faster than 21.59% of Python online submissions for Reverse Prefix of Word.
Memory Usage: 13.3 MB, less than 96.92% of Python online submissions for Reverse Prefix of Word.
原题链接：leetcode.com/problems/re…
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