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 "++dcba++efd".

Return the resulting string.

Example 1:

复制代码
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:

复制代码
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:

复制代码
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:

复制代码
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:

复制代码
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();
    }
}
相关推荐
Navigator_Z6 小时前
LeetCode //C - 1240. Tiling a Rectangle with the Fewest Squares
c语言·算法·leetcode
稻米哟6 小时前
力扣100——双指针
算法·leetcode
橘子汽水16810 小时前
Leetcode 198,118打家劫舍,杨辉三角
算法·leetcode
alphaTao14 小时前
LeetCode 每日一题 2026/9/7-2026/9/13
算法·leetcode
土司大王16 小时前
LeetCode 79 单词搜索:Java 回溯模板、网格 DFS 与剪枝优化
java·算法·leetcode·深度优先
All for pursuit.17 小时前
【数组-5】560.和为K的子数组
数据结构·c++·算法·leetcode
土司大王2 天前
LeetCode 17 电话号码的字母组合:Java 回溯模板、多叉决策树与复杂度分析
java·leetcode·决策树
find1star2 天前
LeetCode 25:K 个一组翻转链表
java·数据结构·算法·leetcode·链表·职场和发展·动态规划