LeetCode解题报告--ZigZag Conversion

题目来源：https://leetcode.com/problems/zigzag-conversion/

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解法一：
间距为: icount = 2 * (nRows - 1)
每列的字符个数为：i = nRows[from 0 to (nRows - 1)]
中间未加粗字符其下标为: mid = j + icount - 2 * i (j : from 0 to s.length() - 1)

public static String convert(String s, int numRows) {        int icount = 2 * (numRows - 1);        String newStr = "";        if(numRows < 2 || s.length() <= 1)            return s;        else{            for(int i = 0;i < numRows;i ++){                for(int j = i;j < s.length();j += icount){                    newStr += s.charAt(j);                    int mid = j + icount - 2 * i;                    if(i != 0 && i != numRows - 1 && mid >= 0 && mid < s.length()){                        newStr += s.charAt(mid);                    }                }            }            return newStr;        }    }

解法二：
将每层看层一个字符串，则共有nRows字符串，最后将其依次输出即可。
public static String convert(String s, int numRows) {

        StringBuffer[] str = new StringBuffer[numRows];        boolean down = true;        int j = 0;        if(s.length() <= 1 || numRows <= 1)            return s;        for(int i = 0;i < str.length;i ++)            str[i] = new StringBuffer();        for(int i = 0;i < s.length();i ++){            str[j].append(s.charAt(i));            if(j == 0)                 down = true;            else if(j == numRows - 1)                down =false;            if(down)                j ++;            else                j --;        }        StringBuffer newStr= new StringBuffer();        for(StringBuffer string : str)            newStr.append(string.toString());        return newStr.toString();    }