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.
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".
Constraints:
1 <= word.length <= 250word 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:
- 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:
- 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();
- }
- }