LeetCode 6 ZigZag Conversion（Z型转换）（String）

翻译

字符串“PAYPALISHIRING”通过一个给定的行数写成如下这种Z型模式：
P A H N
A P L S I I G
Y I R

然后一行一行的读取：“PAHNAPLSIIGYIR”

写代码读入一个字符串并通过给定的行数做这个转换：

string convert(string text, int nRows);

调用convert(“PAYPALISHIRING”, 3)，应该返回”PAHNAPLSIIGYIR”。

原文

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”.

分析

如果还是没明白题目的意思，看下图吧……

public class Solution
{public string Convert(string s, int numRows){if (numRows == 1)return s;StringBuilder strBuilder = new StringBuilder();int lengthOfGroup = 2 * numRows - 2;        // 如上图所示，每组的长度为4     for (int row = 0; row < numRows; row++)      // 按从第0行到numRows-1行的顺序遍历  {if (row == 0 || row == numRows - 1)        // 此处负责第0行和numRows-1行{for (int j = row; j < s.Length; j += lengthOfGroup){strBuilder.Append(s[j]);}}else                   // 此处负责第0行和numRows-1行之间的所有行{int currentRow = row;           // 在当前行中向右移动（看上图）bool flag = true;int childLenOfGroup1 = 2 * (numRows - 1 - row);                //  怎么说呢……中间行的各个索引吧int childLenOfGroup2 = lengthOfGroup - childLenOfGroup1;while (currentRow < s.Length){strBuilder.Append(s[currentRow]);if (flag)currentRow += childLenOfGroup1;elsecurrentRow += childLenOfGroup2;flag = !flag;}}}return strBuilder.ToString();}
}

C++的代码肯定是有的：

class Solution {
public:string convert(string s, int numRows) {      if(numRows==1)return s;string str="";int lengthOfGroup=2*numRows-2;for(int row=0;row<numRows;row++){if(row==0||row==numRows-1){for(int currentRow=row;currentRow<s.length();currentRow+=lengthOfGroup){str+=s[currentRow];}}else{int currentRow=row;bool flag=true;int childLenOfGroup1=2*(numRows-1-row);int childLenOfGroup2=lengthOfGroup-childLenOfGroup1;while(currentRow<s.length()){str+=s[currentRow];if(flag)currentRow+=childLenOfGroup1;elsecurrentRow+=childLenOfGroup2;flag=!flag;}}}return str;   }
};

至于Java嘛，当然也有……不过每次我都是直接把C#的代码拷贝过去然后改改就好了。

public class Solution {public String convert(String s, int numRows) {if (numRows == 1)return s;StringBuilder strBuilder = new StringBuilder();int lengthOfGroup = 2 * numRows - 2;       for(int row=0; row < numRows; row++){if (row == 0 || row == numRows - 1){for(int currentRow = row; currentRow < s.length(); currentRow += lengthOfGroup){strBuilder.append(s.charAt(currentRow));}}else{int currentRow = row;        boolean flag = true;int childLenOfGroup1 = 2 * (numRows - 1 - row);           int childLenOfGroup2 = lengthOfGroup - childLenOfGroup1;while (currentRow <s.length()){strBuilder.append(s.charAt(currentRow));if (flag)currentRow += childLenOfGroup1;elsecurrentRow += childLenOfGroup2;flag = !flag;}}}return strBuilder.toString();}
}

updated at 2016/09/17

做一个小幅修改~

还是Java代码，这里首先将String转成了Char数组，其实不转也一样。while循环中有两个for，一个是如上图中的向下遍历，一个向上遍历，向上遍历的时候也只是取中间段。因为虽然图形上画出来是倾斜的，但是在Char数组中其实还是连续的，所以也不用过多担心。

后面的for循环，其实只是将所有的StringBuilder，集中到第一个然后作为输出而已。

    public String convert(String s, int numRows) {char[] c = s.toCharArray();int len = c.length;StringBuilder[] sb = new StringBuilder[numRows];for (int i = 0; i < sb.length; i++) {sb[i] = new StringBuilder();}int index = 0;while (index < len) {for (int i = 0; i < numRows && index < len; i++) {sb[i].append(c[index++]);}for (int i = numRows - 2; i >= 1 && index < len; i--) {sb[i].append(c[index++]);}}for (int i = 1; i < sb.length; i++) {sb[0].append(sb[i]);}return sb[0].toString();}

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LeetCode 6 ZigZag Conversion（Z型转换）（String）

翻译

原文

分析

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