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"
,
return5
.
分三步:
代码如下:
C++ code int lengthOfLastWord(const char *s) {
int i, j, len = 0;
for (i = 0; s[i] != ‘\0‘; ++i) ;
for ( --i; s[i] == ‘ ‘; --i) ;
for (j = i; j >= 0 && s[j] != ‘ ‘; --j, ++len) ;
return len;
}
一趟找到最后那个单词的长度。
C++ code int lengthOfLastWord(const char *s) {
int len = 0, last = 0;
while (*s)
if (*s++ == ‘ ‘) {
last = len > 0 ? len : last;
len = 0;
}
else ++len;
return len > 0 ? len : last;
}
LeetCode[string]: Length of Last Word
原文:http://blog.csdn.net/chfe007/article/details/40859251