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.

这题是要求最后一个单词的长度。而且因为输入是一个char *，而不是已经长度的数组，所以从左往右扫。碰到空格就保留它的位置。碰到非空格就计算长度。如果前面没有空格，就直接和字符串开头相比较。如果有空格就和最近的空格比较。

这种方法就可以忽略末尾是不是有空格，或者连续空格等情况了。

1 class Solution { 2 public: 3     int lengthOfLastWord(const char *s) { 4         const char *p = s, *space = NULL; 5         int l = 0; 6         for (; *p != '\0'; p++) { 7             if (*p == ' ') { 8                 space = p; 9             } else if (space == NULL) {10                 l = p - s + 1;11             } else {12                 l = p - space;13             }14         }15         return l;16     }17 };