锯齿转变（ZigZag Conversion）

来源：互联网发布：税务数据采集软件编辑：程序博客网时间：2024/04/29 22:52

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   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".

这个题目看完以后我猜大部分人还是没搞明白题目到底想干啥，我当时也没搞明白，主要原因还是例子给的有点少，好，既然他例子给的少，那我们自己再举几个栗子……

假设输入字符串为“ABCDEFGHIJKLMNOPQRSTUVWXYZ”

那么当nRow = 3时：

A E I M Q U Y BDFHJLNPRTVXZ C G K O S W

那么当nRow = 4时：

A  G  M  S  Y  B FH LN RT XZ  CE IK OQ UW    D  J  P  V

那么当nRow = 5时：

A   I   Q   Y   B  HJ  PR  XZ   C G K O S W     DF  LN  TV      E   M   U

这下就很明白了，代码的实现不是很难，只要控制好边界条件即可。我的思路也很简单，按照当前字符串的字符顺序填入二维数组中，然后再去便利二维数组……感觉时间复杂度上还是控制的不太好，以后有了更好的方法再更新吧……

public class ZigZagConversion {        public String convert(String s, int numRows) {        if (numRows == 1)            return s;        int len = s.length();        int times = len / (numRows + numRows - 2);        int tailLen = len % (numRows + numRows - 2);        StringBuilder stringBuilder = new StringBuilder("");        char zzArr[][] = new char[numRows][(times + 1) * (numRows - 1)];        int k = 0;        boolean tag = false;        boolean clo = true;        boolean dig = false;        for (int i = 0; i < (times + 1) * (numRows - 1);){            if (clo){                for (int j = 0; j < numRows;){                    if (k < s.length())                        zzArr[j][i] = s.charAt(k++);                    j++;                    if (j < numRows -1){                        clo = true;                        dig = false;                    }                    if (j == numRows - 1){                        clo = false;                        dig = true;                    }                }            }            if (dig){                for (int j = numRows - 1; j > 0;) {                    j--;                    i++;                    if (j > 0 & k < s.length())                        zzArr[j][i] = s.charAt(k++);                    if (j > 0){                        clo = false;                        dig = true;                    }                    if (j == 0){                        clo = true;                        dig = false;                    }                }            }            if (k == s.length())                break;        }        for (int i = 0; i < zzArr.length; i++){            for (int j = 0; j < zzArr[i].length; j++){                if (zzArr[i][j] >= '!' && zzArr[i][j] <= '~')                    stringBuilder.append(zzArr[i][j]);            }        }        return stringBuilder.toString();    }}

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