leetcode oj s_06

public class Solution {

    public static void main(String[] args) {
        String s = "PAYPALISHIRING";
        String res = convert(s, 4);
        System.out.println(res);
    }

    /**
     * numRows=1和numRows=2为特殊情况
     */
    public static String convert(String s, int numRows) {
        String res = "";
        int l = s.length();
        if (l == 0) {
            return "";
        }

        if (l > 0 && l <= numRows) {
            return s;
        }

        if (numRows == 1) {
            return s;
        }

        // col为列数
        int col = l / (2 * numRows - 2);
        int remainder = l % (2 * numRows - 2);
        if (remainder >= 0 && remainder <= numRows) {
            col = 2 * col + 1;
        }
        if (remainder > numRows) {
            col = 2 * col + 2;
        }

        // temp为辅助数组
        int[] temp = new int[col];
        temp[0] = 1;
        for (int i = 1; i < col; i++) {
            temp[i] = 2 * i * (numRows - 1) - temp[i - 1];
        }
        for (int i = 0; i < numRows; i++) {
            if (i == 0) {
                int j = 0;
                while (2 * j * (numRows - 1) < l) {
                    res += s.charAt(2 * j * (numRows - 1));
                    j++;
                }
                continue;
            }
            if (i == numRows - 1) {
                int j = 0;
                while ((2 * j + 1) * (numRows - 1) < l) {
                    res += s.charAt((2 * j + 1) * (numRows - 1));
                    j++;
                }
                continue;
            }
            for (int k = 0; k < col; k++) {
                if (k == 0 && i < l) {
                    res += s.charAt(i);
                    continue;
                }
                
                if(k%2==0){
                    if(temp[k]+i-1<l){
                        res += s.charAt(temp[k]+i-1);
                        continue;
                    }
                }
                
                if (k % 2 == 1) {
                    if (temp[k] - i + 1 < l) {
                        res += s.charAt(temp[k] - i + 1);
                        continue;
                    }
                }
                break;
            }

        }

        return res;
    }
}

 思路：

另一种解法：

/**
     * 时间复杂度也为O(n)  但更快
     */
    public static String convert1(String s, int numRows) {
        if(numRows==1) return s;
        int x = 2 * (numRows-1); // distance between pipes |/|/|...
        int len = s.length();
        char[] c = new char[len];
        int k =0;
        for(int i=0; i < numRows; i++)
        {
            for(int j=i;j<len;j=j+x)
            {
                c[k++] = s.charAt(j);
                if(i>0 && i<numRows-1 && j+x-2*i < len)
                {
                       c[k++] = s.charAt(j+x-2*i); // extra character between pipes
                }
            }
        }
        return new String(c);
    }

思路比我的更加清晰。通过加断点debug可以理解算法思想。