leetcode 38. Count and Say

leetcode 38. Count and Say  计数和报数（c语言实现）

Description

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" or11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

题目的意思：

n=1，为 1

n=2，表示n=1有一个1，结果就为11

n=3，表示n=2有2个1，结果就为21

n=4，表示n=3有1个2和1个1，结果就为1211

n=5，表示n=4有1个1，1个2，2个1，结果就为111221

n=6，表示n=5有3个1，2个2，1个1，结果就为312211

以此类推。

char* countAndSay(int n) {    char *precountAndSay;    int length, i, j = 0, count = 1;    char * retStr;    if(n == 1)     {        retStr = (char*)malloc(sizeof(char) * 2);//多的一个存'\0'        retStr[j++] = '1';    }    else if(n == 2){        retStr = (char*)malloc(sizeof(char) * 3);        retStr[j++] = '1';        retStr[j++] = '1';    }    else if(n == 3)     {        retStr = (char*)malloc(sizeof(char) * 3);        retStr[j++] = '2';        retStr[j++] = '1';    }    if(n > 3)    {        precountAndSay= countAndSay(n-1);//找到上一次的串，递归法        length = strlen(precountAndSay);//上次串的长度        retStr = (char*)malloc(sizeof(char) * (2 * length + 1));//最差的情况，分配的内存是length的2倍加1        for(i = 1; i < length; i++)        {            if(precountAndSay[i-1] == precountAndSay[i])//相等，数有多少个这样的数，存于count            {                count ++;                continue;            }else{                retStr[j++] = count + '0';//将count转化为1-9的字符                retStr[j++] = precountAndSay[i - 1];//存发现count计数后出现不同时的前一个值                count = 1;            }        }        //处理最后一个字符        retStr[j++] = count + '0';        retStr[j++] = precountAndSay[i - 1];                        }    retStr[j] = '\0';//转成字符串    return retStr;}

这个题目，run有个bug，无论输入多少，官方给的期待的值始终为null。但submit solution是Accepted的，已经反馈了，但不知改了没。