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:

 1     public static String convert(String s, int nRows) {
 2 
 3         Character[][] zigzag = new Character[nRows][s.length()];
 4         String reslut = new String();
 5         int length = 0;
 6         int slength = s.length();
 7         int j;
 8 
 9         if (slength <= 1 || nRows == 1) {
10             return s;
11         }
12 
13         for (j = 0; length < slength; j++) {
14             if (j % (nRows - 1) == 0) {
15                 for (int i = 0; i < nRows && length < slength; i++) {
16                     zigzag[i][j] = s.charAt(length);
17                     length++;
18                 }
19             } else {
20 21                 zigzag[nRows - j % (nRows - 1) - 1][j] = s.charAt(length);
22                 length++;
23 
24             }
25 
26         }
27 
28         for (int i = 0; i < nRows; i++) {
29             for (int k = 0; k <= j; k++) {
30                 if (zigzag[i][k] != null) {
31 32                     reslut = reslut + zigzag[i][k];
33                 }
34             }
35         }
36         return reslut;
37     }

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：

 1     public String convert(String s, int nRows) {
 2 
 3         int length = s.length();
 4         if (length <= 1 || nRows == 1) {
 5             return s;
 6         }
 7 
 8         String reslut = new String();
 9 
10         int i = 0;
11         int j = 0;
12         int[] temp = new int[s.length()];
13         while (j < length) {
14             temp[i] = j;
15             j = j + 2 * (nRows - 1);
16             i++;
17         }
18 
19         int flag = 0;
20         while (flag < nRows) {
21             if (flag == 0) {
22                 int k = 0;
23                 while (k < i) {
24                     reslut = reslut + s.charAt(temp[k]);
25                     k++;
26                 }
27             } else if (flag == nRows - 1) {
28                 int k = 0;
29                 while (k < i) {
30                     if (temp[k] + nRows - 1 < length) {
31                         reslut = reslut + s.charAt(temp[k] + nRows - 1);
32                     }
33                     k++;
34                 }
35             } else {
36                 int k = 0;
37                 while (k < i) {
38 
39                     if (temp[k] + flag < length) {
40                         reslut = reslut + s.charAt(temp[k] + flag);
41                     }
42 
43                     if (temp[k] + 2 * nRows - flag - 2 < length) {
44                         reslut = reslut + s.charAt(temp[k] + 2 * nRows - flag - 2);
45                     }
46 
47                     k++;
48                 }
49             }
50             flag++;
51         }
52         return reslut;
53     }

 

posted @ 2014-02-26 21:16  kcrosswind  阅读(233)  评论(0)  编辑  收藏  举报