【LeetCode】ZigZag Conversion 解题报告

【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".

【解析】

第一次看到这个题目的人,可能不知道ZigZag是什么意思,简单解释一下,就是把字符串原顺序012345……按下图所示排列:

比较直观的解法是,用一个字符串数组 string[rows] 来存储每一行,最后一拼接就是最终结果。

用一个delta表示正向还是反向,即上图中从第一行到最后一行还是最后一行到第一行。

代码如下所示:

public class Solution {
    public String convert(String s, int nRows) {
        int len = s.length();
        if (len == 0 || nRows <= 1) return s;

        String[] ans = new String[nRows];
        Arrays.fill(ans, "");
        int row = 0, delta = 1;
        for (int i = 0; i < len; i++) {
            ans[row] += s.charAt(i);
            row += delta;
            if (row >= nRows) {
                row = nRows-2;
                delta = -1;
            }
            if (row < 0) {
                row = 1;
                delta = 1;
            }
        }

        String ret = "";
        for (int i = 0; i < nRows; i++) {
            ret += ans[i];
        }
        return ret;
    }
}

【网上解法】

如 http://blog.csdn.net/cshaxu/article/details/12507201 说的最为简洁:

发现所有行的重复周期都是 2 * nRows - 2

对于首行和末行之间的行,还会额外重复一次,重复的这一次距离本周期起始字符的距离是 2 * nRows - 2 - 2 * i

代码如下所示:

public class Solution {
    public String convert(String s, int nRows) {
        int len = s.length();
        if (len == 0 || nRows < 2) return s;

        String ret = "";
        int lag = 2*nRows - 2; //循环周期
        for (int i = 0; i < nRows; i++) {
            for (int j = i; j < len; j += lag) {
                ret += s.charAt(j);

                //非首行和末行时还要加一个
                if (i > 0 && i < nRows-1) {
                    int t = j + lag - 2*i;
                    if (t < len) {
                        ret += s.charAt(t);
                    }
                }
            }
        }
        return ret;
    }
}

【总结】

这道题属于简单题,找规律即可,循环周期可能比较容易找,周期中间的规律 2 * nRows - 2 - 2 * i 可能不大好找。

时间: 2024-11-05 23:27:05

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