[LeetCode] 6. ZigZag Conversion

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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 s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

题目：输入一个字符串和一个目标行数，按照目标行数返回这个字符串的“锯齿”模式（见上面的样例）。

实现思路：按照样例的形式把字符串s写成k行的锯齿模式，实现过程中先求出每一行的字符集合，然后把所有行的字符集合拼在一起就是最终结果。从下图不难发现，在顺序遍历所有字符时，按照“向下、向上、向下、向上...”的方向即可求出每一行的字符集合。用布尔值goDown控制方向，goDown = true表示向下走（行数递增），goDown = false表示向上走（行数递减），当row=1或row=k时用goDown = !goDown改变方向。这种方法的好处就是非常直观，不需要锯齿模式的字符规律。（找规律的方法可以参考下面的链接）

class Solution {
    public String convert(String s, int numRows) {
        if (s == null) throw new IllegalArgumentException("argument is null");
        if (numRows == 0) throw new IllegalArgumentException("unexpected number of rows");
        if (numRows == 1) return s;
        List<StringBuilder> rows = new ArrayList<>();
        boolean goDown = false;
        int row = 0;
        StringBuilder result = new StringBuilder();
        for (int i = 0; i < Math.min(numRows, s.length()); i++)
            rows.add(new StringBuilder());
        for (char c : s.toCharArray()) {
            rows.get(row).append(c);
            if (row == 0 || row == numRows - 1)
                goDown = !goDown;
            row = goDown ? row + 1 : row - 1;
        }
        for (StringBuilder sb : rows)
            result.append(sb);
        return result.toString();
    }
}

参考资料：

https://leetcode.com/problems/zigzag-conversion/solution/

https://leetcode.com/problems/zigzag-conversion/discuss/3435/If-you-are-confused-with-zigzag-patterncome-and-see!