题目：ZigZag Conversion

一串字符按照Z字形的数组给了我们，要求转成原本的顺序。

思路：

统计“|/”的个数；

竖着的和斜着的下标有对应关系；

竖着的：k = j*(2*numRows - 2) + i;

斜着的(不含两个端点)：k = (j + 1)*(2*numRows - 2) - i;

注意：可能会有残缺的部分。

/******************************************************************ZigZag ConversionThe string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)P   A   H   NA P L S I I GY   I   RAnd then read line by line: "PAHNAPLSIIGYIR"Write the code that will take a string and make this conversion given a number of rows:string convert(string text, int nRows);convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".******************************************************************/#include 
     
      #include 
      
       char* convert(char* s, int numRows) {    int length = strlen(s);    printf("%d\n",length);    if(length <= numRows || numRows <= 1)return s;    int zn = length/(2*numRows - 2);//Z字形竖着放时，最开始的一竖和一撇的组合整体的个数    int sn = length%(2*numRows - 2);//Z字形竖着放时，剩下残缺的组合的字母个数    char *cs = (char *)malloc((length + 1)*sizeof(char));    memset(cs,0,(length + 1)*sizeof(char));    int index = 0,k = 0;    for (int i = 0;i < numRows;i++){
   //完整组合的对应转换        for(int j = 0;j < zn;j++){
   //竖线上的点坐标对应公式            k = j*(2*numRows - 2) + i;            cs[index++] = s[k];            if(i > 0 && i < numRows - 1){
   //斜线上的不含两端点的点坐标对应公式                k = (j + 1)*(2*numRows - 2) - i;                cs[index++] = s[k];            }        }        if(sn > i){
   //残缺组合的对应转换            k = zn*(2*numRows - 2) + i;            cs[index++] = s[k];            if(i > 0 && i < numRows - 1 && sn > 2*numRows - 2 - i){                k = (zn + 1)*(2*numRows - 2) - i;                cs[index++] = s[k];            }        }    }    return cs;}void main(){    char s[] = "Apalindromeisaword,phrase,number,orothersequenceofunitsthatcanbereadthesamewayineitherdirection,withgeneralallowancesforadjustmentstopunctuationandworddividers.";    char *cs = convert(s,2);    printf("%s\n",cs);    free(cs);}