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

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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……

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

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;
                    else
                        currentRow += 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;
                    else
                        currentRow+=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;
                      else
                          currentRow += 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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