PAT- 字符串压缩与解压

https://pintia.cn/problem-sets/994805260223102976/problems/994805262018265088

#include<iostream>
using namespace std;
int main()
{
	char n;
	scanf("%c",&n);
	getchar();	//读掉回车
	string s;
	getline(cin,s);
	int count,i;
	if (n=='C') {	//压缩
		count=1;
		for (i=0; i<s.length(); i++) {
			if (s[i]==s[i+1]) {	//当前字符跟后面的一样 
				count++;
			} else {
				if (count>1)	//大于1需要加数字
					printf("%d",count);
				printf("%c",s[i]);
				count=1;
			}
		}
	} else if (n=='D') {	//解压
		for (i=0; i<s.length(); i++) {
			if (s[i]>='0' && s[i]<='9') {	//如果当前是数字
				if (s[i+1]>='0' && s[i+1]<='9') { //下一个还是数字
					count = count*10+s[i]-'0';
					continue;	//转到下一个字符位置
				} else {
					count = count*10+s[i++]-'0';
				}
			}
			count= count==0?1:count;	//count是0的话就赋值1
			while(count>0) {	
				printf("%c",s[i]);
				count--;
			}	//最后count一定会变为0
		}
	}
	return 0;
}