557.Reverse Words in a String III(String-Easy)

转载请注明作者和出处： http://blog.csdn.net/c406495762

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

Input: "Let's take LeetCode contest"Output: "s'teL ekat edoCteeL tsetnoc"

Note: In the string, each word is separated by single space and there will not be any extra space in the string.

题目：反转字符串。给定一个字符串，保留字符串中空格和每个单词的顺序，反转每个单词。

思路：以空格为标志，对每个小结的单词进行操作

Language : cpp

class Solution {public:    string reverseWords(string s) {        int front = 0;                  //记录空格后第一个字符的位置        int str_length = s.length();    //字符串长度        for(int i = 0; i <= str_length; i++) {            if(i == s.length() || s[i] == ' ') {    //找到空格反转，没有空格整个字符串反转                reverse(&s[front], &s[i]);                front = i + 1;                     }        }        return s;    }};

Language : python

    第一种最笨的方法：先使用split以空格为标志分开字符串，然后对每个字符串进行反转操作，再将字符串进行拼接。

class Solution(object):    def reverseWords(self, s):        """        :type s: str        :rtype: str        """        result = ''        input_str = s.split(' ')        for each in input_str:            result = result + each[::-1] + ' '        return result[:-1]

    第二种方法，对于第一种方法进行了改进，使用join进行拼接。

class Solution(object):    def reverseWords(self, s):        """        :type s: str        :rtype: str        """        return ' '.join(x[::-1] for x in s.split())

    第三种方法，不实用迭代方法，加快运行速度。首先反转每个单词的顺序，然后反转整个字符串：

class Solution(object):    def reverseWords(self, s):        """        :type s: str        :rtype: str        """        return ' '.join(s.split()[::-1])[::-1]

    我的LeetCode代码获取：https://github.com/Jack-Cherish/LeetCode