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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".
自己一上来没搞懂什么是zigzag，还以为就是多了个中点，结果怎么都不对，搞得相当复杂，权且贴上代码，当个教训吧：
1 public class ZigzagConversion { 2 public static String convert(String s, int numRows) { 3 if (s == null || s.length() == 0) 4 return s; 5 if (numRows < 2) 6 return s; 7 int gap = numRows - 2; 8 String res = ""; 9 10 for (int i = 0; i < numRows; i++) { 11 if (i == 0) { 12 for (int j = i; j < s.length(); j += (numRows * 2 - 2)) 13 res += s.charAt(j); 14 } else if (i == numRows - 1) { 15 for (int j = i; j < s.length(); j += (numRows * 2 - 2)) 16 res += s.charAt(j); 17 } else { 18 for (int j = i; j < s.length(); j += (numRows * 2 - 2)) { 19 res += s.charAt(j); 20 if (j + numRows * 2 - 3 - i < s.length()) 21 res += s.charAt(j + numRows * 2 - 3 - i); 22 } 23 } 24 } 25 26 return res; 27 } 28 29 public static void main(String[] args) { 30 String s = "ABCDE"; 31 String res = convert(s, 4); 32 System.out.println(res); 33 } 34 35 }
问题代码，边界问题没有解决好
1 public class Solution { 2 public String convert(String s, int nRows) { 3 if (s == null || s.length() <= nRows || nRows <= 1) return s; 4 StringBuffer sb = new StringBuffer(); 5 // the first row 6 for (int i = 0; i < s.length(); i += (nRows - 1) * 2) { 7 sb.append(s.charAt(i)); 8 } 9 10 for (int i = 1; i < nRows - 1; i++) { 11 for (int j = i; j < s.length(); j+= (nRows - 1) * 2) { 12 sb.append(s.charAt(j)); 13 if (j + (nRows - i - 1) * 2 < s.length()) { 14 sb.append(s.charAt(j + (nRows - i - 1) * 2)); 15 } 16 } 17 } 18 // the last row 19 for (int i = nRows - 1; i < s.length(); i += (nRows - 1) * 2) { 20 sb.append(s.charAt(i)); 21 } 22 return sb.toString(); 23 } 24 }
后来百度了下，看到别人代码相当简洁，方知自己压根就没搞懂题目。
Zigzag:即循环对角线结构（
| 0 | 8 | 16 | |||||||||
| 1 | 7 | 9 | 15 | 17 | |||||||
| 2 | 6 | 10 | 14 | 18 | |||||||
| 3 | 5 | 11 | 13 | 19 | |||||||
| 4 | 12 | 20 |
）
构建n个字符串，循环为每个分别添加相应的字符，最后合并所有字符；
自己照着思路重新写了下：
1 public class ZigzagConversion { 2 public static String convert(String s, int numRows) { 3 if (s == null || s.length() == 0) 4 return s; 5 if (numRows < 2) 6 return s; 7 String[] res = new String[numRows]; 8 String result = ""; 9 10 int i = 0, j; 11 12 while (i < s.length()) { 13 for (j = 0; j < numRows && i < s.length(); j++) { 14 res[j] += s.charAt(i); 15 i++; 16 } 17 for (j = numRows - 2; j > 0 && i < s.length(); j--) { 18 res[j] += s.charAt(i); 19 i++; 20 } 21 } 22 for(int k = 0; k < numRows;k++) 23 result += res[k]; 24 return result; 25 } 26 27 public static void main(String[] args) { 28 String s = "PAYPALISHIRING"; 29 String res = convert(s, 3); 30 System.out.println(res); 31 } 32 }
只是没次字符串数组都默认初始化为null，如何覆盖还没解决，原文用C++代码如下：
1 string convert(string s, int nRows){ 2 if(nRows == 1) return s; 3 string res[nRows]; 4 int i = 0, j, gap = nRows-2; 5 while(i < s.size()){ 6 for(j = 0; i < s.size() && j < nRows; ++j) res[j] += s[i++]; 7 for(j = gap; i < s.size() && j > 0; --j) res[j] += s[i++]; 8 } 9 string str = ""; 10 for(i = 0; i < nRows; ++i) 11 str += res[i]; 12 return str; 13 }
发现自己一些问题：首先是不能很好的思考问题，或者说自己的思路多少总会有些偏差；其次就是边界问题，总是犯些小错误。
LeetCode 6 -- ZigZag Conversion
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原文地址：http://www.cnblogs.com/myshuangwaiwai/p/4483004.html
