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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.
这道题真是费了九牛二虎之力，本来没有多难的题目，自己本来对这道题感觉确实挺简单的。但是很多情况没有考虑周全，导致代码频繁出现bug。最后终于调试通过了
我的代码：
class Solution { public: int lengthOfLastWord(const char *s) { if (strlen(s) == 0) return 0; int length = strlen(s); int count = 0; while (length - count - 1>=0) { if ((s[length - count - 1] >= ‘a‘ && s[length - count - 1] <= ‘z‘ || s[length - count - 1] >= ‘A‘&&s[length - count - 1] <= ‘Z‘) && length - count - 1 >= 0) { count++; if (length - count == 0 || s[length - count - 1] == ‘ ‘) break; } if (s[length - count - 1] == ‘ ‘) { length --; continue; } } return count; } };
在参考别人的代码，真是感觉自愧不如啊。下面的这种解法，又简单又直观
int lengthOfLastWord(const char* s) { int len = 0; while (*s) { if (*s++ != ‘ ‘) ++len; else if (*s && *s != ‘ ‘) len = 0; } return len; }
HappyLeetcode35:Length of Last Word
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原文地址：http://www.cnblogs.com/chengxuyuanxiaowang/p/4189233.html
