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"
andch = "d"
, then you should reverse the segment that starts at0
and ends at3
(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".
Constraints:
1 <= word.length <= 250
-
word
consists of lowercase English letters. -
ch
is a lowercase English letter.
就很简单,先找到string里char的位置,然后把这个位置及其之前的string反过来就行了,最后return一个新string。
纯自己写:文章来源:https://www.toymoban.com/news/detail-702206.html
class Solution {
public String reversePrefix(String word, char ch) {
StringBuilder sb = new StringBuilder();
int index = -1;
for (int i = 0; i < word.length(); i++) {
if (word.charAt(i) == ch) {
index = i;
break;
}
}
if (index == -1) {
return word;
}
for (int i = index; i >= 0; i--) {
sb.append(word.charAt(i));
}
for (int i = index + 1; i < word.length(); i++) {
sb.append(word.charAt(i));
}
return sb.toString();
}
}
看了下solutions还可以纯用java api,巧妙的用了StringBuilder的substring()和reverse()方法:文章来源地址https://www.toymoban.com/news/detail-702206.html
class Solution {
public String reversePrefix(String word, char ch) {
int index = word.indexOf(ch);
if (index == -1) {
return word;
}
StringBuilder sb = new StringBuilder(word.substring(0, index + 1)).reverse();
return sb.append(word.substring(index + 1, word.length())).toString();
}
}
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