/*
1. 暴力解法： 都投影到y轴，用指针判断投影顺序并输出，是否奇形怪状的题大部分都用这种方法？
2. 找规律：因为是循环的，所以可以找规律，
        等差数列: https://www.acwing.com/solution/LeetCode/content/75/

*/


class Solution {
    public String convert(String s, int numRows) {
        if (s == null || s.length() == 0 || s.length() == 1 || numRows == 0 || numRows == 1) return s;
        var res = arithSeq(s, numRows);
        return res;
    }

    public String bruteForce(String s, int n){
        List<StringBuilder> list = new ArrayList<>();
        for (int i = 0 ; i < n ; i++) list.add(new StringBuilder());

        int next = -1;
        int index = 0;
        for (int i = 0; i < s.length(); i++){
            char c = s.charAt(i);

            list.get(index).append(c);
            if (index % n == 0 || index % n == n-1) next = -next;
            // 会导致index 读到 n， 因为在第一个循环节后 i==n 时 index == n-2 ， 如果用i%n则 i%n == 0 导致再次转下向
            // 故更新判断应该用最近的一个变量
            // if (i % n == 0 || i % n == n-1) next = -next;  

            index += next;

        }

        var buffer = list.get(0);
        for (int i = 1 ; i < n ;i ++){
            buffer.append(list.get(i).toString());
        }

        return buffer.toString();

    }


    public String arithSeq(String s, int n){
        StringBuilder buffer = new StringBuilder();
        int step = 2*n-2;
        for (int i = 0 ; i < n ;i ++){
            for (int k = i, j = step - i; k < s.length() || j < s.length() ; k += step, j+= step){
                if (k < s.length()) buffer.append(s.charAt(k));
                if (i == 0 || i == n-1) continue;
                if (j < s.length()) buffer.append(s.charAt(j));
            }
        }
        return buffer.toString();
    }
}