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【日常训练--腾讯精选50】557. 反转字符串中的单词 III
2022-07-05 08:36:00 【Puppet__】
题目
给定一个字符串 s ,你需要反转字符串中每个单词的字符顺序,同时仍保留空格和单词的初始顺序。
示例 1:
输入:s = “Let’s take LeetCode contest”
输出:“s’teL ekat edoCteeL tsetnoc”
示例 2:
输入: s = “God Ding”
输出:“doG gniD”
提示:
1 <= s.length <= 5 * 104
s 包含可打印的 ASCII 字符。
s 不包含任何开头或结尾空格。
s 里 至少 有一个词。
s 中的所有单词都用一个空格隔开
代码
package tencent50;
public class leetcode557 {
// 使用额外空间,并通过原字符串中的空格将切分为多段,每段进行翻转
public String reverseWords(String s) {
StringBuffer sb = new StringBuffer();
int len = s.length();
int i = 0;
while( i < len){
int start = i;
// 先找空格在哪里
while (i < len && s.charAt(i) != ' '){
i++;
}
// 将单词翻转
for (int j = i - 1; j >= start; j--){
sb.append(s.charAt(j));
}
//跳过之后的空格
while (i < len && s.charAt(i) == ' '){
i++;
sb.append(' ');
}
}
return sb.toString();
}
public static void main(String[] args) {
leetcode557 obj = new leetcode557();
System.out.println(obj.reverseWords("Let's take LeetCode contest"));
}
}
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