2128. Reverse Prefix of Word

难度: Easy

题目描述

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".

 

Constraints:

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

解题思路

代码实现

解决方案

java

class Solution {
    public String reversePrefix(String word, char ch) {
        LinkedList<Character> stack = new LinkedList<>();
        char[] charArray = word.toCharArray();
        StringBuilder sb = new StringBuilder();
        int i = 0;
        for (char c : charArray) {
            if (c == ch) {
                stack.push(c);

                break;
            }
            stack.push(c);
            i++;
        }
        if(i ==charArray.length){
            return word;
        }
        while (!stack.isEmpty()) {
            sb.append(stack.pop());
        }

        for (int j = i + 1; j < charArray.length; j++) {
            sb.append(charArray[j]);
        }

        return sb.toString();
    }
}