ZigZag Conversion 把一个字符串转为zigzag的表示形式@LeetCode

我发现有一类题目，比如这一道题，都是属于思路好想，但是很容易编程有bug的题目。

要特别细心处理corner case。。。

这里还看到一个比较不容易出错的方法，就是把nRows个string叠加起来，然后连接起来就好  http://blog.csdn.net/errant_xia/article/details/8639920

package Level3;import java.util.Arrays;/** * 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   NA P L S I I GY   I   RAnd 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".Discuss */public class S6 {public static void main(String[] args) {//System.out.println(convert("PAYPALISHIRING", 4));System.out.println(convert("ABCD", 2));}public static String convert(String s, int nRows) {if(s.length() == 0){return s;}int nCols = s.length();char[][] buf = new char[nRows][nCols];for(char[] ch : buf){Arrays.fill(ch, ' ');}int x = 0, y = 0;boolean down = true;for(int i=0; i<s.length();){buf[x][y] = s.charAt(i);if(down) {// 往下走if(x == nRows-1){// 走到最底，改变方向if(x > 0){// 注意当只有一行时可能出现负数！x--;}y++;down = false;}else{// 往下走x++;}i++;}else{// 往右上走if(x == 0){// 仅仅改变方向，不前进idown = true;}if(x >=1){// 继续往右上方走x--;y++;i++;}}}String ret = "";for(int i=0; i<nRows; i++){for(int j=0; j<nCols; j++){if(buf[i][j] != ' '){ret +=  buf[i][j];}}}return ret;}}

Again:

public static String convert(String s, int nRows) {if(nRows<=1 || s.length()<2){return s;}String[] ss = new String[nRows];// 用来存储每一行Arrays.fill(ss, "");int nGroup = 2 * (nRows-1);// 每nGroup，zigzag pattern就会重复一遍for(int i=0; i<s.length(); i++){// 推导出的公式/* eg: nRows=4, nGroup=6 i%group: 0,1,2,3,4,5 nRows-1-(i%nGroup): 3,2,1,0,-1,-2 abs():3,2,1,0,1,2 total:0,1,2,3,2,1 */ss[nRows-1-Math.abs(nRows-1-(i%nGroup))] += s.charAt(i);}String ret = "";for(String str : ss){ret += str;}return ret;}

两个规律：

1 两个zigzag之间间距为2*nRows-2

2 每个zigzag中间（在j和j+interval之间）位置为j+interval-2*i

public String convert(String s, int nRows) {    int len = s.length();    int interval = 2 * nRows - 2;   // interval for zigzag    if(len<=1 || nRows<=1) {        return s;    }    String ret = "";    for(int i=0; i<nRows; i++) {        // row-wise        for(int j=i; j<len; j+=interval) {  // next zigzag            ret += s.charAt(j);            int mid = j + interval - 2 * i; // middle value in zigzag            // ignore first and last row            if(i!=0 && i!=nRows-1 && mid >= 0 && mid < len) {                ret += s.charAt(mid);            }        }    }    return ret;}