给定一句英文，除了字母之外，还包含空格回车和水平制表符号('\t', '\n'), 根据这三个符号来计算一句英文中所含有单词的个数。
下面给出的这个方法是我从一个国外网站上看到的，思路清晰而且很有逻辑性，于是决定记录下来：
设定两个标志: IN, OUT和一个变量state，当遇到字母的时候state值为IN， 当遇到上面说的那三个字符的时候state为OUT。你可以测试任何情况，包括两个单词中间有多个空格的情况，下面给出代码：
[cpp]
#include
#include
using namespace std;
unsigned int count_word(char *s) {
const int OUT = 0;
const int IN = 1;
int state = OUT;
unsigned int count = 0;
while (*s) {
if (*s == ' ' || *s == '\t' || *s == '\n')
state = OUT;
else if (OUT == state) {
state = IN;
++count;
}
s++;
}
return count;
}
int main(int argc, char *argv[]) {
char s[] = "this is a test\n this is a test";
cout << count_word(s) << endl;
cin.get();
return 0;
}
#include
#include
using namespace std;
unsigned int count_word(char *s) {
const int OUT = 0;
const int IN = 1;
int state = OUT;
unsigned int count = 0;
while (*s) {
if (*s == ' ' || *s == '\t' || *s == '\n')
state = OUT;
else if (OUT == state) {
state = IN;
++count;
}
s++;
}
return count;
}
int main(int argc, char *argv[]) {
char s[] = "this is a test\n this is a test";
cout << count_word(s) << endl;
cin.get();
return 0;
}