字符串按照Z旋转90度然后上下翻转的字形按行输出字符串--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 s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I
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个人思路:

先计算出列数,用二维数组处理

然后分为字符落在竖和斜杠两种情况

竖:分为是否可以落满行的个数

斜杠同理

同时计算出当前竖的第一个值的列号

    public String convert(String s, int numRows) {//ZigZag Conversion
        int len=s.length();
        if(numRows==1||s==null||s.equals(" ")||len<=numRows) return s;
        else{
             char[] result=new char[len];
             char[] cs=s.toCharArray();
             int cLen = 0;
             int split1=2*(numRows-1);
             int nDouble=len/split1;
             if(len%split1==0){
                 cLen=(1+numRows-2)*nDouble;

             }else if(len%split1<=numRows){
                 cLen=(1+numRows-2)*nDouble+1;
             }else{
                 cLen=(1+numRows-2)*nDouble+(len%split1)-(numRows-1);
             }
             
             char[][] ch=new char[numRows][cLen];
             int dif=split1;
             for(int k=0;k<len;k++){
//                System.out.println("start++++::::"+k);
                 if(k%dif==0){  //up start
                     int t=k>0?(k/2):0;
                    if(k+numRows>len){
                        for(int j=0;j<len-k;j++){
//                            System.out.println("45-----"+j+","+t+"="+(k+j));
                            ch[j][t]=cs[k+j];
                        }
                        k=len-1;
                        break;
                    }else{
                    for(int j=0;j<numRows;j++){
//                        System.out.println("52-----"+j+","+t+"="+(k+j));
                        ch[j][t]=cs[k+j];
                    }
                    k=k+numRows-2;
                    }
                    
                 }else if((k%dif==(numRows-1))){//down start
//                    System.out.println(k);
                    int t=k+1-numRows>0?((k+1-numRows)/2):0;
                    int tt=numRows-1;
                     if(k+numRows-2>=len){
                        for(int j=0;j<len-k-1;j++){
                               --tt;
                            ++t;
//                            System.out.println("64-----"+tt+","+t+"="+cs[k+j+1]);
                            ch[tt][t]=cs[k+j+1];
                        }
                        k=len-1;
                        break;
                    }else{
                    for(int j=0;j<numRows-2;j++){
                        
                        --tt;
                        ++t;
//                        System.out.println("t:"+tt);
//                        System.out.println("70-----"+tt+","+t+"="+cs[k+j+1]);
                        ch[tt][t]=cs[k+j+1];
                    }
                    k=k+numRows-2;
//                    System.out.println("79----"+k);
                    }
                    
                 }
             }
             int next=0;
             for(int i=0;i<ch.length;i++){
                 for(int j=0;j<ch[0].length;j++){
//                     System.out.print(ch[i][j]);
                     if(ch[i][j]!='\u0000'){
                         result[next++]=ch[i][j];
                     }
                 }
             }
             
             return new String(result);
        }
        
 
    }

 

posted on 2019-04-14 00:46  Honey_Badger  阅读(...)  评论(...编辑  收藏

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