[LeetCode] Reverse Words in a String II 翻转字符串中的单词之二
Given an input string , reverse the string word by word.
Example:
Input: ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]
Note:
- A word is defined as a sequence of non-space characters.
- The input string does not contain leading or trailing spaces.
- The words are always separated by a single space.
Follow up: Could you do it in-place without allocating extra space?
这道题让我们翻转一个字符串中的单词,跟之前那题 Reverse Words in a String 没有区别,由于之前那道题就是用 in-place 的方法做的,而这道题反而更简化了题目,因为不考虑首尾空格了和单词之间的多空格了,方法还是很简单,先把每个单词翻转一遍,再把整个字符串翻转一遍,或者也可以调换个顺序,先翻转整个字符串,再翻转每个单词,参见代码如下:
解法一:
class Solution {
public:
void reverseWords(vector<char>& str) {
int left = , n = str.size();
for (int i = ; i <= n; ++i) {
if (i == n || str[i] == ' ') {
reverse(str, left, i - );
left = i + ;
}
}
reverse(str, , n - );
}
void reverse(vector<char>& str, int left, int right) {
while (left < right) {
char t = str[left];
str[left] = str[right];
str[right] = t;
++left; --right;
}
}
};
我们也可以使用 C++ STL 中自带的 reverse 函数来做,先把整个字符串翻转一下,然后再来扫描每个字符,用两个指针,一个指向开头,另一个开始遍历,遇到空格停止,这样两个指针之间就确定了一个单词的范围,直接调用 reverse 函数翻转,然后移动头指针到下一个位置,在用另一个指针继续扫描,重复上述步骤即可,参见代码如下:
解法二:
class Solution {
public:
void reverseWords(vector<char>& str) {
reverse(str.begin(), str.end());
for (int i = , j = ; i < str.size(); i = j + ) {
for (j = i; j < str.size(); ++j) {
if (str[j] == ' ') break;
}
reverse(str.begin() + i, str.begin() + j);
}
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/186
类似题目:
参考资料:
https://leetcode.com/problems/reverse-words-in-a-string-ii/
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] Reverse Words in a String II 翻转字符串中的单词之二的更多相关文章
- [LeetCode] 186. Reverse Words in a String II 翻转字符串中的单词 II
Given an input string, reverse the string word by word. A word is defined as a sequence of non-space ...
- [LeetCode] Reverse Words in a String III 翻转字符串中的单词之三
Given a string, you need to reverse the order of characters in each word within a sentence while sti ...
- [LeetCode] 557. Reverse Words in a String III 翻转字符串中的单词 III
Given a string, you need to reverse the order of characters in each word within a sentence while sti ...
- LeetCode刷题:Reverse Words in a String(翻转字符串中的单词)
题目 Given an input string, reverse the string word by word. For example, Given s = "the sky is b ...
- 186. Reverse Words in a String II 翻转有空格的单词串 里面不变
[抄题]: Given an input string , reverse the string word by word. Example: Input: ["t"," ...
- LeetCode 557. Reverse Words in a String III (反转字符串中的单词 III)
Given a string, you need to reverse the order of characters in each word within a sentence while sti ...
- [LeetCode] 151. Reverse Words in a String 翻转字符串中的单词
Given an input string, reverse the string word by word. For example,Given s = "the sky is blue& ...
- [LeetCode] Reverse Words in a String 翻转字符串中的单词
Given an input string, reverse the string word by word. For example, Given s = "the sky is blue ...
- [LeetCode] Reverse String II 翻转字符串之二
Given a string and an integer k, you need to reverse the first k characters for every 2k characters ...
随机推荐
- 3.JAVA之GUI编程Frame窗口
创建图形化界面思路: 1.创建frame窗体: 2.对窗体进行基本设置: 比如大小.位置.布局 3.定义组件: 4.将组件通过add方法添加到窗体中: 5.让窗体显示,通过setVisible(tur ...
- Android来电监听和去电监听
我觉得写文章就得写得有用一些的,必须要有自己的思想,关于来电去电监听将按照下面三个问题展开 1.监听来电去电有什么用? 2.怎么监听,来电去电监听方式一样吗? 3.实战,有什么需要特别注意地方? 监听 ...
- Scala化规则引擎
1. 引言 什么是规则引擎 一个业务规则包含一组条件和在此条件下执行的操作,它们表示业务规则应用程序的一段业务逻辑.业务规则通常应该由业务分析人员和策略管理者开发和修改,但有些复杂的业务规则也可以由技 ...
- TFS 2013 培训视频
最近给某企业培训了完整的 TFS 2013 系列课程,一共四天. 下面是该课程的内容安排: 项目管理 建立项目 成员的维护 Backlog 定义 任务拆分 迭代 ...
- HTML基础标签
[HTML写法标签][HTML字体段落标签][锚点][有序无序列表][表格] 一.HTML写法标签:双标签:<标签名>内容</标签名>单标签:<标签名 内容/> 二 ...
- Gradle project sync failed
在Android Studio中运行APP时出现了以下错误: gradle project sync failed. please fix your project and try again 解决的 ...
- MYSQL 开发技巧
主要涉及:JOIN .JOIN 更新.GROUP BY HAVING 数据查重/去重 1 INNER JOIN.LEFT JOIN.RIGHT JOIN.FULL JOIN(MySQL 不支持).CR ...
- 《Web开发过滤Javascript、HTML的方法》
JavaScript过滤方法: 第一种方案:使用 htmlspecialchars 函数转换特殊字符和使用 nl2br 函数插入一些必要的 <br /> 标签. $comment = &l ...
- Unicode简介
计算机只能处理二进制,因此需要把文字表示为二进制才能被计算机理解和识别. 一般的做法是为每一个字母或汉字分配一个id,然后用二进制表示这个id,存在内存或磁盘中.计算机可以根据二进制数据知道这个id是 ...
- css揭秘--笔记(未完)
第0章 关于本书 1, 本书要用到一个工具函数————$$(),它可以让我们更容易获取和遍历所有匹配特定css选择符的dom元素: function $$(selector,context){ con ...