ZigZag Conversion2015年6月23日

题目：

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".

思路：

1、通过字符串的长度和numRows大小算出商和余数

2、通过商和余数判断每一行元素的个数，各行元素个数存放在int[] row

3、逐行对应出每一行元素在源字符串中的位置，并根据位置取出相应字符添加到返回字符串的末尾

解答：

 /  test cases passed.
Status: Accepted
Runtime:  ms
Submitted:  minutes ago

public class Solution {
     public String convert(String s, int numRows) {
            StringBuilder result = new StringBuilder("");
            int len = s.length();
            if(numRows == 1 || len<=numRows){
                return s;
            }
            int quotient = len / (2*numRows -2);
            int remainder = len % (2*numRows -2);
            int[] row = new int[numRows];

            //求数组row
            for(int i=0; i<numRows; i++){
                if(i==0){
                    if(remainder > i) {
                        row[i] = quotient + 1;
                    }else{
                        row[i] = quotient;
                    }

                }else if(i == numRows -1){
                    if(remainder > i) {
                        row[i] = quotient + 1;
                    }else{
                        row[i] = quotient;
                    }
                }else{
                    if(remainder > i && remainder <= numRows) {
                        row[i] = 2*quotient + 1;
                    }else if(remainder > numRows){
                        if((remainder - numRows) >= (numRows-1-i)){
                             row[i] = 2*quotient + 2;
                        }else{
                             row[i] = 2*quotient + 1;
                        }

                    }else{
                        row[i] = 2*quotient;
                    }
                }
            }

            //逐行取出源字符串对应位置的字符添加到result
            for(int i=0; i<numRows; i++) {
                if(i==0 || i==numRows-1){
                    for(int j=0; j<row[i]; j++ ){
                         result.append(s.charAt(i+j*(2*numRows-2)));
                    }
                }else{
                    for(int j=0; j<row[i]; j++ ){
                        if(j%2==0){
                            result.append(s.charAt(i+j/2*(2*numRows-2)));
                        }else{
                            result.append(s.charAt(i+(j-1)/2*(2*numRows-2)+2*numRows-2-2*i));
                        }

                    }
                }

            }

            return result.toString();

      }
}