标签：i++ xpl paypal nat 个数 col break inpu array
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
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); } }
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
标签：i++ xpl paypal nat 个数 col break inpu array
原文地址：https://www.cnblogs.com/tk55/p/10703768.html
