leetcode 6. 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”.

这个题的就是说，有一个字符数字是按照Z子形的形式构造的，然后请你正确的读出正确的数组。 解决方法很简单，就是按照它的要求构造一遍，然后就可以读出正确的结果。
代码如下：

/* * 这道题就是把Z子形的字符串正确的读出来， * 思路很简单就是按照Z子形的方式读一遍， * 然后在取出结果就可以了 *  * */public class Solution {     public String convert(String s, int numRows)     {         if(s==null || s.length()<=0 || numRows<=1)             return s;         //这里把结果保存在string字符数组内         StringBuilder[] builders=new StringBuilder[numRows];         for(int i=0;i<numRows;i++)             builders[i]=new StringBuilder();         //通过row来判断Z子形的走向和步进         int row=0,step=1;         for(int i=0;i<s.length();i++)         {             builders[row].append(s.charAt(i));             row+=step;             if(row==numRows)             {                 row=row-2;                 step=-1;             }else if(row==-1)             {                 row=row+2;                 step=1;             }         }         String res="";         for(int i=0;i<builders.length;i++)             res+=builders[i].toString();         return res;     }    public static void main(String[] args)     {        Solution solution=new Solution();        System.out.println(solution.convert("ABC", 2));    }}

下面是C++版本的答案，就是按照要求构造一遍，然后直接读去一遍即可完成。

代码如下：

#include <iostream>#include <string>#include <vector>using namespace std;class Solution {public:    string convert(string s, int numRows)     {        if (numRows <= 1)            return s;        string res = "";        vector<string> zip;        for (int i = 0; i < numRows; i++)            zip.push_back("");        int row = 0,step = 1;        for (int i = 0; i < s.length(); i++)        {            zip[row] += s[i];            row += step;            if (row == numRows)            {                row -= 2;                step = -1;            }            else if (row == -1)            {                row += 2;                step = 1;            }        }        for (int i = 0; i < numRows; i++)            res += zip[i];        return res;    }};