Reverse Words in a String 反转单词顺序

Reverse Words in a String

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

For example, Given s = "the sky is blue", return "blue is sky the".

Update (2015-02-12): For C programmers: Try to solve it in-place in O(1) space

使用API 复杂度

时间 O(N) 空间 O(N)

思路

将单词根据空格split开来存入一个字符串数组，然后将该数组反转即可。

注意

先用trim()将前后无用的空格去掉

用正则表达式" +"来匹配一个或多个空格

代码

1. public class Solution {
2. public String reverseWords(String s) {
3. String[] words = s.trim().split(" +");
4. int len = words.length;
5. StringBuilder result = new StringBuilder();
6. for(int i = len -1; i>=0;i--){
7. result.append(words[i]);
8. if(i!=0) result.append(" ");
9. }
10. return result.toString();
11. }
12. }

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双指针交换法 复杂度

时间 O(N) 空间 O(N) 如果输入时char数组则是O(1)

思路

先将字符串转换成char的数组，然后将整个数组反转。然后我们再对每一个单词单独的反转一次，方法是用两个指针记录当前单词的起始位置和终止位置，遇到空格就进入下一个单词。

代码

1. public class Solution {
2. public String reverseWords(String s) {
3. s = s.trim();
4. char[] str = s.toCharArray();
5. // 先反转整个数组
6. reverse(str, 0, str.length - 1);
7. int start = 0, end = 0;
8. for(int i = 0; i < s.length(); i++){
9. if(str[i]!=' '){
10. end++;
11. } else {
12. // 反转每个单词
13. reverse(str, start, end - 1);
14. end++;
15. start = end;
16. }
17. }
18. //反转最后一个单词，因为其后面没有空串

19. reverse(str,start,str.length-1);

    
    
20. return String.valueOf(str);
21. }
23. public void reverse(char[] str, int start, int end){
24. while(start < end){
25. char tmp = str[start];
26. str[start] = str[end];
27. str[end] = tmp;
28. start++;
29. end--;
30. }
31. }
32. }

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Reverse Words in a String II

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.

The input string does not contain leading or trailing spaces and the words are always separated by a single space.

For example, Given s = "the sky is blue", return "blue is sky the". 
Could you do it in-place without allocating extra space?

双指针交换法 复杂度

时间 O(N) 空间 O(1)

思路

这题就是Java版的Inplace做法了，先反转整个数组，再对每个词反转。

代码

1. public class Solution {
2. public void reverseWords(char[] s) {
3. reverse(s, 0, s.length - 1);
4. int start = 0;
5. for(int i = 0; i < s.length; i++){
6. if(s[i] == ' '){
7. reverse(s, start, i - 1);
8. start = i + 1;
9. }
10. }
11. reverse(s, start, s.length - 1);
12. }
14. public void reverse(char[] s, int start, int end){
15. while(start < end){
16. char tmp = s[start];
17. s[start] = s[end];
18. s[end] = tmp;
19. start++;
20. end--;
21. }
22. }
23. }