翻譯
<code class="hljs" mathematica="">字符串“PAYPALISHIRING”通過(guò)一個(gè)給定的行數(shù)寫(xiě)成如下這種Z型模式：P A H NA P L S I I GY I R然后一行一行的讀取：“PAHNAPLSIIGYIR”寫(xiě)代碼讀入一個(gè)字符串并通過(guò)給定的行數(shù)做這個(gè)轉(zhuǎn)換：string convert(string text, int nRows);調(diào)用convert(PAYPALISHIRING, 3)，應(yīng)該返回PAHNAPLSIIGYIR，LeetCode 6 ZigZag Conversion（Z型轉(zhuǎn)換）
。</code>
原文
<code applescript="" class="hljs">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 NA P L S I I GY I RAnd then read line by line: PAHNAPLSIIGYIRWrite 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.</code>
如果還是沒(méi)明白題目的意思，看下圖吧……
<code class="hljs" cs="">public class Solution{ public string Convert(string s, int numRows) { if (numRows == 1) return s; StringBuilder strBuilder = new StringBuilder(); int lengthOfGroup = 2 * numRows - 2; // 如上圖所示，每組的長(zhǎng)度為4 for (int row = 0; row < numRows; row++) // 按從第0行到numRows-1行的順序遍歷 { if (row == 0 || row == numRows - 1) // 此處負(fù)責(zé)第0行和numRows-1行 { for (int j = row; j < s.Length; j += lengthOfGroup) { strBuilder.Append(s[j]); } } else // 此處負(fù)責(zé)第0行和numRows-1行之間的所有行 { int currentRow = row; // 在當(dāng)前行中向右移動(dòng)（看上圖） bool flag = true; int childLenOfGroup1 = 2 * (numRows - 1 - row); // 怎么說(shuō)呢……中間行的各個(gè)索引吧 int childLenOfGroup2 = lengthOfGroup - childLenOfGroup1; while (currentRow < s.Length) { strBuilder.Append(s[currentRow]); if (flag) currentRow += childLenOfGroup1; else currentRow += childLenOfGroup2; flag = !flag; } } } return strBuilder.ToString(); }}</code>C++的代碼肯定是有的：
<code class="hljs" cs="">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++){ bool="" childlenofgroup1="2*(numRows-1-row);" childlenofgroup2="lengthOfGroup-childLenOfGroup1;" code="" currentrow="row;currentRow<s.length();currentRow+=lengthOfGroup){" else="" flag="true;" int="" return="" row="=0||row==numRows-1){"></numrows;row++){></code> 至于Java嘛，當(dāng)然也有……不過(guò)每次我都是直接把C#的代碼拷貝過(guò)去然后改改就好了，電腦資料
《LeetCode 6 ZigZag Conversion（Z型轉(zhuǎn)換）》(http://www.ipr-jzsc.com)。
<code class="hljs" cs=""><code class="hljs" cs="">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()){ code="" currentrow="" else="" flag="!flag;" if="" return=""></s.length()){></code></code>