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, 

求字符串最后一个单词的长度

1. 之前考虑不全,　当只用一个length时,会出现'a '时判断错误,　于是要用两个标志来指明

2. 其实也可以先求长度,　从后往前就是求第一个单词的长度

class Solution {public:    int lengthOfLastWord(const char *s) {        int length1  = 0;        int length2  = 0;        int i,j;        char c;        while(*s!='\0')        {            if(*s++!=' ')                 length1++;            else if(length1!=0)            {                length2 = length1;                 length1 = 0;            }        }        if(length1!=0)            return length1;        if(length1==0)            return length2;    }};

这个是从尾到头,　直接求第一个单词长度,这样就是最后一个的长度了

有几点:

１.　const char *p如果不加const就会报错,　类型不一样

２.　s + lenght - 1; 这里要减１

class Solution {public:    int lengthOfLastWord(const char *s) {       int length = strlen(s);       int i;       int len = 0;       const char *p = s  + length - 1;       while(length--)       {        if(*p!=' ')        {            len++;            p--;        }        else if(len!=0)                break;            else                {                    p--;                    continue;                }       }        return len;    }};