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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
.
先把字符串最后的空格去掉,然后统计从后往前第一个单词中字母的个数。
1 int lengthOfLastWord(char *s) { 2 if(s == NULL) return 0; 3 int i = strlen(s)-1; 4 int count = 0; 5 while(s[i] == ‘ ‘) i--; 6 while(s[i] != ‘ ‘ && i>=0){ 7 count++; 8 i--; 9 } 10 return count; 11 }
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原文地址:http://www.cnblogs.com/zhhc/p/4353776.html