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。