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【日常訓練--騰訊精選50】557. 反轉字符串中的單詞 III
2022-07-05 08:37:00 【Puppet__】
題目
給定一個字符串 s ,你需要反轉字符串中每個單詞的字符順序,同時仍保留空格和單詞的初始順序。
示例 1:
輸入:s = “Let’s take LeetCode contest”
輸出:“s’teL ekat edoCteeL tsetnoc”
示例 2:
輸入: s = “God Ding”
輸出:“doG gniD”
提示:
1 <= s.length <= 5 * 104
s 包含可打印的 ASCII 字符。
s 不包含任何開頭或結尾空格。
s 裏 至少 有一個詞。
s 中的所有單詞都用一個空格隔開
代碼
package tencent50;
public class leetcode557 {
// 使用額外空間,並通過原字符串中的空格將切分為多段,每段進行翻轉
public String reverseWords(String s) {
StringBuffer sb = new StringBuffer();
int len = s.length();
int i = 0;
while( i < len){
int start = i;
// 先找空格在哪裏
while (i < len && s.charAt(i) != ' '){
i++;
}
// 將單詞翻轉
for (int j = i - 1; j >= start; j--){
sb.append(s.charAt(j));
}
//跳過之後的空格
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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