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"

 

Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

[解题思路]

第0, len-1行各个字符之间的间隔为2×nRows - 2

第1~len-2行之间会有两个固定间隔之间会有其他字符插入

规律：index + 2 * nRows - 2 * i - 2, 这里i是指行号

 

 1 public class Solution {
 2     public static String convert(String s, int nRows) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if (s == null || s.length() == 0 || nRows <= 1) {
 6             return s;
 7         }
 8 
 9         String result = generate(s, nRows);
10         return result;
11     }
12 
13     private String generate(String s, int nRows) {
14         int len = s.length();
15         StringBuffer buffer = new StringBuffer();
16         int diff = 2 * nRows - 2;
17         for (int i = 0; i < nRows && i < len; i++) {
18             if (i == 0 || i == nRows - 1) {
19                 buffer.append(s.charAt(i));
20                 int index = i;
21                 while (index + diff < len) {
22                     buffer.append(s.charAt(index + diff));
23                     index = index + diff;
24                 }
25 //                int tmp = i;
26 //                while (i + diff < len) {
27 //                    buffer.append(s.charAt(i + diff));
28 //                    i = i + diff;
29 //                }
30 //                i = tmp;
31             } else {
32                 buffer.append(s.charAt(i));
33 //                int tmp = i;
34                 int index = i;
35                 while (2 * nRows - 2 * i - 2 + index < len
36                         || index + diff < len) {
37                     if(2 * nRows - 2 * i - 2 + index < len){
38                         buffer.append(s.charAt(2 * nRows - 2 * i - 2 + index));
39                     }
40                     if (index + diff < len) {
41                         buffer.append(s.charAt(index + diff));
42                     }
43                     index += diff;
44                 }
45 //                while (2 * nRows - i - 2 < len || i + diff < len) {
46 //                    if (2 * nRows - i - 2 < len && 2 * nRows - i - 2 >= 0) {
47 //                        buffer.append(s.charAt(2 * nRows - i - 2));
48 //                    }
49 //                    if (i + diff < len) {
50 //                        buffer.append(s.charAt(i + diff));
51 //                    }
52 //                    i = i + diff;
53 //                }
54 //                i = tmp;
55             }
56         }
57         return buffer.toString();
58     }
59 }

易错点：

1.i 是行号，不是字符串的index

2.特殊情况没有考虑：nRows == 1时，应当直接返回输入, 最初没有考虑==1的情况，出现死循环，导致Memory Limit Exceeded

3.当nRows > len 时，没有考虑，可能会出现下标溢出的情况

posted @ 2013-09-06 11:36  feiling  阅读(717)  评论(0)  编辑  收藏  举报