LEET code -- Length of Last Word

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.




但是這解法不好
因為strlen本身就會對s 字串跑一次  太浪費

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int lengthOfLastWord(char* s) { int size=strlen(s); if(size<=0)return 0; int count=0; for(int i=size-1;i>=0;i--){ if(count==0 && s[i]==' ')continue; else if(s[i]!=' ')count++; else if(s[i]==' ')return count; } return count; }

strlen

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size_t strlen(str) const char *str; { register const char *s; for (s = str; *s; ++s); return(s - str); }

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跑一次for就好

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int lengthOfLastWord(char* s) { //int size=strlen(s); //if(size<=0)return 0; int count=0,lastcount=0; for(;*s;s++){ if(count==0 && *s==' ')continue; else if(*s!=' ')count++; else if(*s==' '){ lastcount = count; count=0; } } return count!=0 ? count:lastcount; }