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length of last word
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class Solution { | |
public: | |
int lengthOfLastWord(const char *s) { | |
int len=0; | |
/* 字符串的遍历方式,始终这样写,下面肯定要有个s++ */ | |
while(*s!='\0'){ | |
/* 当本字符不是空格的时候,+1 */ | |
if(*s!=' ') | |
len++; | |
/* 如果本字符等于空格,而下一个字符不等于空格的时候,重新开始计数 */ | |
if(*s++==' '&& *s!=' ' && *s!='\0') | |
/* 一种特殊情况需考虑,空格后面马上是结束符,那个时候也不应该清零,还是前面的长度 */ | |
len=0; | |
/* 连续多个空格的时候,还保持上一次的len */ | |
} | |
return len; | |
/* 如果输入为空, 直接返回0 | |
如果输入为空格,直接返回0 */ | |
} | |
}; |
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