leetcode-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"

不难，但是要注意时间按复杂度和“zigzag”的定义。

zigzag的含义：。应该这样排列字符

根据题意可以暴力来解，新建个二维数组来存对象，再输出。如下:

    public static String convert(String s, int nRows) {

        Character[][] zigzag = new Character[nRows][s.length()];
        String reslut = new String();
        int length = 0;
        int slength = s.length();
        int j;

        if (slength <= 1 || nRows == 1) {
            return s;
        }

        for (j = 0; length < slength; j++) {
            if (j % (nRows - 1) == 0) {
                for (int i = 0; i < nRows && length < slength; i++) {
                    zigzag[i][j] = s.charAt(length);
                    length++;
                }
            } else {
21                 zigzag[nRows - j % (nRows - 1) - 1][j] = s.charAt(length);
                length++;

            }

        }

        for (int i = 0; i < nRows; i++) {
            for (int k = 0; k <= j; k++) {
                if (zigzag[i][k] != null) {
32                     reslut = reslut + zigzag[i][k];
                }
            }
        }
        return reslut;
    }

问题是会超时。

其实这个题只要找到规律直接扫描就可以了：

这确的代码：

    public String convert(String s, int nRows) {

        int length = s.length();
        if (length <= 1 || nRows == 1) {
            return s;
        }

        String reslut = new String();

        int i = 0;
        int j = 0;
        int[] temp = new int[s.length()];
        while (j < length) {
            temp[i] = j;
            j = j + 2 * (nRows - 1);
            i++;
        }

        int flag = 0;
        while (flag < nRows) {
            if (flag == 0) {
                int k = 0;
                while (k < i) {
                    reslut = reslut + s.charAt(temp[k]);
                    k++;
                }
            } else if (flag == nRows - 1) {
                int k = 0;
                while (k < i) {
                    if (temp[k] + nRows - 1 < length) {
                        reslut = reslut + s.charAt(temp[k] + nRows - 1);
                    }
                    k++;
                }
            } else {
                int k = 0;
                while (k < i) {

                    if (temp[k] + flag < length) {
                        reslut = reslut + s.charAt(temp[k] + flag);
                    }

                    if (temp[k] + 2 * nRows - flag - 2 < length) {
                        reslut = reslut + s.charAt(temp[k] + 2 * nRows - flag - 2);
                    }

                    k++;
                }
            }
            flag++;
        }
        return reslut;
    }