LeetCode:58 Length of Last Word

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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;     } }


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