[LEETCODE] Length of Last Word

[LEETCODE]  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.

A word is defined as a character sequence consists of non-space characters only.

For example,
Given s = "Hello World",
return 5.

思路:

最后一个单词的2种情况：

1， *hello; 2， *hello* ; *代表空格，可以为一个也可以为多个;
从前向后扫描，遇到空格就计算当前单词的长度，然后移动指针到下一个不是空格的地方继续计算，如果是情况1，指针移动到最后的‘\0’ 时，长度还未计算，所以循环结束之后再计算一次长度。

代码:

int lengthOfLastWord(const char *s) {
int len =0;
const char *start = s;
while( *s !='\0'){
if( *s == ' '){
len = s - start;
while( *s == ' ' ){ s++; }
start = s;
continue;
}
s++;
}
if( *start != '\0')
len = s -  start;
return len;
}