leetcode第38题-Count and Say

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the nth sequence.

Note: The sequence of integers will be represented as a string.

题目要求：主要是数数，但是题目要求是输出第n个字符串，即如果输入5，则输出字符串111221。

分析：按照题意，可以首先定义两个字符串，一个用来表示当前cur，一个用来表示进阶到的下一个tmp，然后遍历时每遇到相同的就使count++，然后就用'0'+count表示这个数有几个。

完整代码:

#include<stdio.h>#include<stdlib.h>#include<string.h>char* countAndSay(int n) {    if(n==1)return "1";char *cur=(char*)malloc(sizeof(char)*2),*tmp;cur[0]='1',cur[1]=0;int len,index,i,j,count;for(i=2;i<=n;i++){len=strlen(cur);tmp=(char*)malloc(sizeof(char)*(3*len));memset(tmp,0,3*len);count=1;for(index=1,j=0;index<len;index++){if(cur[index]==cur[index-1])count++;else{tmp[j++]='0'+count;tmp[j++]=cur[index-1];count=1;}}tmp[j++]='0'+count;tmp[j++]=cur[len-1];free(cur);cur=tmp;}return cur;}int main(){    int n;    while(scanf("%d",&n)!=EOF){printf("%s\n",countAndSay(n));    }    return 0;}