zigzag convert

/**zigzag covnert*/#include <stdio.h>#include <string.h>#include <stdlib.h>/*n=numRowsΔ=2n-2    1                           2n-1                         4n-3Δ=        2                     2n-2  2n                    4n-4   4n-2Δ=        3               2n-3        2n+1              4n-5       .Δ=        .           .               .               .            .Δ=        .       n+2                 .           3n               .Δ=        n-1 n+1                     3n-3    3n-1                 5n-5Δ=2n-2    n                           3n-2                         5n-4That is the zigzag pattern the question asked.注意每一行两个数据间的步进,setp1 setp2.*/char* convert(char *s, int numRows){if(numRows==1)return s;int s_len=strlen(s);char *s_convert=(char*)malloc((s_len+1)*sizeof(char));char *s_r=s_convert;int i=0;int step1,step2;int pos;for(;i<numRows;++i){step1=(numRows-i-1)<<1;step2=i<<1;pos=i;if(pos<s_len) *s_convert++=s[pos];while(1){pos+=step1;if(pos>=s_len)break;if(step1)*s_convert++=s[pos];pos+=step2;if(pos>=s_len)break;if(step2)*s_convert++=s[pos];}}*s_convert='\0';return s_r;}int main(int argc,char **argv){char s[]="123456789";printf("%s\n",convert(s,2));return 0;}