[LeetCode] 2000. Reverse Prefix of Word

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".

Constraints:
1 <= word.length <= 250
word consists of lowercase English letters.
ch is a lowercase English letter.

反转单词前缀。

给你一个下标从 0 开始的字符串 word 和一个字符 ch 。找出 ch 第一次出现的下标 i ，反转 word 中从下标 0 开始、直到下标 i 结束（含下标 i ）的那段字符。如果 word 中不存在字符 ch ，则无需进行任何操作。

例如，如果 word = "abcdefd" 且 ch = "d" ，那么你应该 反转 从下标 0 开始、直到下标 3 结束（含下标 3 ）。结果字符串将会是 "dcbaefd" 。
返回 结果字符串 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-prefix-of-word
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思路

题意不难理解，反转 input 字符串的某一个前缀，返回反转之后的结果。需要考虑的 corner case 是如果字符串中不存在目标字母，则返回原字符串。一般的 case 是如果找到了目标字母第一次出现的位置 i，则对这个前缀 (0, i) 进行反转，与字符串剩余的部分拼接好之后返回即可。

复杂度

时间O(n)
空间O(n) - StringBuilder

代码

Java实现

class Solution {
    public String reversePrefix(String word, char ch) {
        int i = word.indexOf(ch);
        String a = reverse(word, 0, i);
        return a;
    }

    private String reverse(String word, int start, int end) {
        char[] letters = word.toCharArray();
        while (start < end) {
            char temp = letters[start];
            letters[start] = letters[end];
            letters[end] = temp;
            start++;
            end--;
        }
        return String.valueOf(letters);
    }
}

再提供一个第一次写的像X一样的代码
Java实现

class Solution {
    public String reversePrefix(String word, char ch) {
        // corner case
        if (!word.contains(ch + "")) {
            return word;
        }

        // normal case
        StringBuilder sb = new StringBuilder();
        int i = 0;
        while (i < word.length()) {
            if (word.charAt(i) == ch) {
                sb.append(helper(word, 0, i));
                break;
            }
            i++;
        }
        i++;

        while (i < word.length()) {
            sb.append(word.charAt(i));
            i++;
        }
        return sb.toString();
    }

    private String helper(String word, int start, int end) {
        char[] w = word.substring(start, end + 1).toCharArray();
        while (start < end) {
            char temp = w[start];
            w[start] = w[end];
            w[end] = temp;
            start++;
            end--;
        }
        StringBuilder sb = new StringBuilder();
        for (char c : w) {
            sb.append(c);
        }
        return sb.toString();
    }
}