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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.
求最后一个非空单词的长度而已，没什么注意点，直接看代码，理解题目意思就可以了：
1 class Solution { 2 public: 3 int lengthOfLastWord(string s) { 4 int len = s.length(); 5 if(len == 0) return 0; 6 while(s[len - 1] == ‘ ‘) 7 if(len - 1 >= 0) 8 len--; 9 else 10 return 0; 11 len--; 12 int lenCount = 0; 13 while(s[len] != ‘ ‘ && len >= 0){ 14 lenCount++, len--; 15 } 16 return lenCount; 17 } 18 };
LeetCode OJ：Length of Last Word（最后一个词的长度）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4900010.html
