LeetCode:58 Length of Last Word

标签：tco   world   case   exist   提交   def   str   define   tac   

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;
     }
}

LeetCode:58 Length of Last Word

标签：tco   world   case   exist   提交   def   str   define   tac   

原文地址：http://www.cnblogs.com/ysmintor/p/7479767.html