ZigZag Conversion

最新推荐文章于 2024-10-24 22:31:46 发布

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本文介绍了一种将字符串转换为指定行数的zigzag模式的方法，并提供了相应的代码实现。

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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：用一个字符串数组来存储每一行的字符串，按照“之字型”遍历字符串的时候，更新每一行的字符串的内容。

public class Solution {
    public String convert(String s, int numRows) {
        StringBuilder[] strs = new StringBuilder[numRows];
        int n = s.length();
        if (numRows <= 0) return null;
        if (n == 0) return "";
        if(numRows==1) return s;
        for (int i = 0; i < numRows; i++)
            strs[i] = new StringBuilder("");
        StringBuilder res = new StringBuilder("");

        for (int i = 0; i < n;) {
            for (int j = 0; i < n && j <= numRows-1; j++) {
                strs[j] = strs[j].append(s.charAt(i));
                i++;
            }
            for (int j = numRows - 2; i < n && j > 0; j--) {
                strs[j] = strs[j].append(s.charAt(i));
                i++;
            }

        }
        for (int k = 0; k < strs.length; k++)
            res = res.append(strs[k]);
        return res.toString();
    }
}