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  • LeetCode:58 Length of Last Word

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