[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...)
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