[LeetCode]ZigZag Conversion

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 Conversion
The 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   N
A P L S I I G
Y   I   R
And 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 <stdio.h>
#include <memory.h>

char* convert(char* s, int numRows) {
    int length = strlen(s);
    printf("%d
",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
",cs);
    free(cs);
}