Given a0-indexedstringwordand a characterch,reversethe segment ofwordthat starts at index0and ends at the index of thefirst occurrenceofch(inclusive). If the characterchdoes not exist inword, do nothing.

• For example, ifword = "abcdefd"andch = "d", then you shouldreversethe segment that starts at0and ends at3
  (inclusive). The resulting string will be"dcbaefd".

Returnthe 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 = "xyxz
 
xe", 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
• wordconsists of lowercase English letters.
• chis a lowercase English letter.

反轉單詞字首。

給你一個下標從 0 開始的字串 word 和一個字元 ch 。找出 ch 第一次出現的下標 i ，反轉 word 中從下標 0 開始、直到下標 i 結束（含下標 i ）的那段字元。如果 word 中不存在字元 ch ，則無需進行任何操作。

例如，如果 word = "abcdefd" 且 ch = "d" ，那麼你應該 反轉 從下標 0 開始、直到下標 3 結束（含下標 3 ）。結果字串將會是 "dcbaefd" 。
返回 結果字串 。

來源：力扣（LeetCode）
連結：https://leetcode-cn.com/problems/reverse-prefix-of-word
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題意不難理解，反轉 input 字串的某一個字首，返回反轉之後的結果。需要考慮的 corner case 是如果字串中不存在目標字母，則返回原字串。一般的 case 是如果找到了目標字母第一次出現的位置 i，則對這個字首 (0, i) 進行反轉，與字串剩餘的部分拼接好之後返回即可。

時間O(n)

空間O(n) - StringBuilder

Java實現

 1 class Solution {
 2     public String reversePrefix(String word, char ch) {
 3         // corner case
 4         if (!word.contains(ch + "")) {
 5             return word;
 6         }
 7 
 8         // normal case
 9         StringBuilder sb = new StringBuilder();
10         int i = 0;
11         while (i < word.length()) {
12             if (word.charAt(i) == ch) {
13                 sb.append(helper(word, 0, i));
14                 break;
15             }
16             i++;
17         }
18         i++;
19 
20         while (i < word.length()) {
21             sb.append(word.charAt(i));
22             i++;
23         }
24         return sb.toString();
25     }
26 
27     private String helper(String word, int start, int end) {
28         char[] w = word.substring(start, end + 1).toCharArray();
29         while (start < end) {
30             char temp = w[start];
31             w[start] = w[end];
32             w[end] = temp;
33             start++;
34             end--;
35         }
36         StringBuilder sb = new StringBuilder();
37         for (char c : w) {
38             sb.append(c);
39         }
40         return sb.toString();
41     }
42 }

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