Leetcode 算法题--ReverseWordsInString

翻转字符串，想到什么写什么。。。我的做法是先trim掉空格，然后从字符串尾部开始扫描，遇到空格则认为一个单词结束，然后copy这个单词。需要注意的地方在于当扫描到最后一个单词的第一个字母时（譬如the sky is blue的t字母），注意单词长度的自增逻辑。

网上还有人的做法是反转整个字符串，然后逐个翻转单词。

 1 package edu.hust.sse.Problems;
 2
 3 //Given s = "the sky is blue",
 4 //return "blue is sky the".
 5 /***
 6  * author @Ivan July 16th,2014
 7  * @param s
 8  * @return
 9  */
10 public class ReverseWordsInString {
11     public static void main(String[] args){
12         ReverseWordsInString rw = new ReverseWordsInString();
13         rw.reverseWords("  abc test   teacher  ");
14     }
15
16     public String reverseWords(String s) {
17         s  = s.trim();
18         if(s.length() == 0){
19             return "";
20         }
21          byte[] a = s.getBytes();
22         byte[] b = new byte[a.length];
23
24         int startPos = 0;
25         int wordLen =0;
26         int bStart  = 0;
27         boolean addSpace = false;
28         for(int i= a.length- 1;i>=0;i--){
29
30             if(a[i] == 32 ){
31                 //copy a word and add a space
32                 if(wordLen >0){
33
34                     //Need add space from the second word
35                     if(addSpace){
36                         b[bStart] = 32;
37                         bStart ++;
38                     }
39
40                     if(i == 0){
41                         startPos  = i;
42                     }else{
43                         startPos = i+1;
44                     }
45                     for(int j=startPos;j<=i+wordLen;j++){
46                         b[bStart] = a[j];
47                         bStart ++;
48                     }
49
50                     //set initial parameter for the next word
51                     addSpace = true;
52                     wordLen = 0;
53                 }
54
55             }
56             //need to continue to scan a complete word
57             else{
58                 wordLen ++;
59                 if(i==0){
60                     //Need add space from the second word
61                     if(addSpace){
62                         b[bStart] = 32;
63                         bStart ++;
64                     }
65
66                     for(int j=0;j<i+wordLen;j++){
67                         b[bStart] = a[j];
68                         bStart ++;
69                     }
70                 }
71             }
72         }
73
74 //        System.out.println(new String(b));
75         return new String(b).trim();
76
77     }
78  }

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