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.

要求是很容易的，但是在提交后才发现处理的情况很多，字面意思可能理解还不太到位。例如对于“a ”,和” a”，希望的输出都是1，这就考验算法对于边界处理的考虑了。

Solution:

class Solution {

    public int lengthOfLastWord(String s) {

        // you have use trim to tackle the case " a" and "a " whose output should be 1 other than 0

        String trimed = s.trim();

        if (trimed.length() == 0) {

             return 0;

        }

        int count = 0;

        for (int lastIndex = trimed.length() - 1; lastIndex >= 0; lastIndex--) {

            char c = trimed.charAt(lastIndex);

             if (Character.isSpaceChar(c))

                break;

            count++;

        }

        return count;

    }

}