151. Reverse Words in a String #

题目 #

Given an input string, reverse the string word by word.

Example 1:

Input: "the sky is blue"
Output: "blue is sky the"

Example 2:

Input: "  hello world!  "
Output: "world! hello"
Explanation: Your reversed string should not contain leading or trailing spaces.

Example 3:

Input: "a good   example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.

Note:

• A word is defined as a sequence of non-space characters.
• Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
• You need to reduce multiple spaces between two words to a single space in the reversed string.

Follow up:

For C programmers, try to solve it in-place in O(1) extra space.

题目大意 #

给定一个字符串，逐个翻转字符串中的每个单词。

说明：

• 无空格字符构成一个单词。
• 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。
• 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。  

进阶：

• 请选用 C 语言的用户尝试使用 O(1) 额外空间复杂度的原地解法。

解题思路 #

• 给出一个中间有空格分隔的字符串，要求把这个字符串按照单词的维度前后翻转。
• 依照题意，先把字符串按照空格分隔成每个小单词，然后把单词前后翻转，最后再把每个单词中间添加空格。

代码 #

package leetcode

import "strings"

func reverseWords151(s string) string {
	ss := strings.Fields(s)
	reverse151(&ss, 0, len(ss)-1)
	return strings.Join(ss, " ")
}

func reverse151(m *[]string, i int, j int) {
	for i <= j {
		(*m)[i], (*m)[j] = (*m)[j], (*m)[i]
		i++
		j--
	}
}