LeetCode 38 Count and Say(C,C++,Java,Python)

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Problem:

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.

Solution:

依次查询每一个字符出现的次数，并统计下来，然后在结果中存入即可，题目数据量比较小。

题目大意：

根据上一个字符串的序列，来统计每个字符出现的次数和什么字符，比如第一个是1，那第二个就是对第一个字符串的描述：1个1=11

Java源代码（208ms）：

public class Solution {    public String countAndSay(int n) {        char[] seq=new char[100000];        char[] bak=new char[100000];        char[] tmp;        char t;        int top=1,index,l,r,num;        seq[0]='1';seq[1]=0;        while(--n >0){            index=0;            for(int i=0;i<top;i++){                num=1;                while(i+1<top && seq[i+1]==seq[i]){i++;num++;}                l=index;                while(num>0){                    bak[index++]=(char)(num%10+'0');                    num/=10;                }                r=index-1;                while(l<r){t=bak[l];bak[l]=bak[r];bak[r]=t;l++;r--;}                bak[index++]=seq[i];            }            top=index;            tmp=seq;seq=bak;bak=tmp;        }        return new String(seq,0,top);    }}

C语言源代码（0ms）：

char* countAndSay(int n) {    char* seq=(char*)malloc(sizeof(char)*100000);    char* bak=(char*)malloc(sizeof(char)*100000);    char* tmp;    char t;    int top=1,i,index,num,l,r;    seq[0]='1';seq[1]=0;    while(--n){        index=0;        for(i=0;i<top;i++){            num=1;            while(i+1<top && seq[i+1]==seq[i]){                i++;                num++;            }            l=index;            while(num>0){                bak[index++]=num%10+'0';                num/=10;            }            r=index-1;            while(l<r){                t=bak[l];bak[l]=bak[r];bak[r]=t;l++;r--;            }            bak[index++]=seq[i];        }        bak[index]=0;        top=index;        tmp=seq;seq=bak;bak=tmp;    }    free(bak);    return seq;}

C++源代码（0ms）：

class Solution {public:    string countAndSay(int n) {        char* seq=(char*)malloc(sizeof(char)*100000);        char* bak=(char*)malloc(sizeof(char)*100000);        char t,*tmp;        int l,r,index,top=1,i,num;        seq[0]='1';seq[1]=0;        while(--n){            index=0;            for(i=0;i<top;i++){                num=1;                while(i+1<top && seq[i+1]==seq[i]){i++;num++;}                l=index;                while(num>0){                    bak[index++]=num%10+'0';                    num/=10;                }                r=index-1;                while(l<r){t=bak[l];bak[l]=bak[r];bak[r]=t;l++;r--;}                bak[index++]=seq[i];            }            bak[index]=0;            top=index;            tmp=seq;seq=bak;bak=tmp;        }        return string(seq);    }};

Python源代码（64ms）数据量小时可用，数据量大的话参照上面三种：

class Solution:    # @param {integer} n    # @return {string}    def countAndSay(self, n):        seq=['1'];top=1;        while n-1>0:            n-=1;index=0;bak=[]            i=0            while i<top:                num=1                while i+1<top and seq[i+1]==seq[i]:i+=1;num+=1                bak.append(chr(num+ord('0')))                bak.append(seq[i])                i+=1            seq=bak;top=len(bak)        return ''.join(seq)

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