题目：
Given a string s consists of upper/lower-case alphabets and empty space characters ‘ ‘, return the length of last word in the string.
If the last word does not exist, return 0.
Note: A word is defined as a character sequence consists of non-space characters only.
For example,
Given s = "Hello World",
return 5.
题解：
这道题主要是考虑一下最后是不是空格，方法是倒着找不是空格的字符并计数，如果遇到空格且计数不是0，说明最后一个单词已经被计数了，所以可以返回了。
代码如下：
1 public int lengthOfLastWord(String s) {
2 if (s == null || s.length() == 0)
3 return 0;
4
5 int len = s.length();
6 int count = 0;
7 for (int i = len - 1; i >= 0; i--) {
8 if (s.charAt(i) != ‘ ‘) {
9 count++;
10 }
11 if(s.charAt(i)==‘ ‘&&count != 0){
12 return count;
13 }
14 }
15 return count;
16 }
2 if (s == null || s.length() == 0)
3 return 0;
4
5 int len = s.length();
6 int count = 0;
7 for (int i = len - 1; i >= 0; i--) {
8 if (s.charAt(i) != ‘ ‘) {
9 count++;
10 }
11 if(s.charAt(i)==‘ ‘&&count != 0){
12 return count;
13 }
14 }
15 return count;
16 }
当然这道题也能用投机取巧的方法，用split函数把字符串按照空格分隔好，返回最后那个就行。。。
代码如下：
1 public int lengthOfLastWord(String s) {
2 String[] a = s.split(" ");
3 if(a == null || a.length == 0)
4 return 0;
5
6 return a[a.length-1].length();
7 }
2 String[] a = s.split(" ");
3 if(a == null || a.length == 0)
4 return 0;
5
6 return a[a.length-1].length();
7 }