代码实现：

char* convert(char *s, int numRows) {
    int len = strlen(s);
    if (len == 0 || numRows == 1) {
        return s;
    }
    char *a = (char*)malloc(sizeof(char) * (len + 1));
    a[len] = '\0';
    int k = 2 * numRows - 2;
    int n = 0;
    for (int i = 0; i < numRows; i++) {
        for (int j = 0; j < len; j++) {
            if (j % k == i || j % k == k - i) {
                a[n++] = s[j];
            }
        }
    }
    return a;
}