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[LeetCode]6. ZigZag ConversionZ

[LeetCode]6. ZigZag ConversionZ

作者: jchen104 | 来源:发表于2018-12-26 21:35 被阅读0次

    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

    题目说的是Z形变换,但是加上是 |/|这样的形状,按Z看了好久一点规律都没有,才发现不是Z形的

    这里找到一个很坑的规律
    对于任意整数序列K[0,1,2,3,4,...]和numRows有:
    1.第0行,索引位置为k(2numRows-2),如0(23-2),1(23-2)等
    2.最后一行,索引位置为k(2numRows-2)+numRows-1,如04+3-1,14+3-1等
    3.其余第i行,索引位置为k(2numRows-2)+i以及(k+1)(2*numRows-2)-i处
    很繁琐的一个规律,有了这个就很好写了

    class Solution {
        public String convert(String s, int numRows) {
    
            if (numRows == 1) return s;
    
            StringBuilder ret = new StringBuilder();
            int n = s.length();
            int cycleLen = 2 * numRows - 2;
    
            for (int i = 0; i < numRows; i++) {
                for (int j = 0; j + i < n; j += cycleLen) {
                    ret.append(s.charAt(j + i));
                    if (i != 0 && i != numRows - 1 && j + cycleLen - i < n)
                        ret.append(s.charAt(j + cycleLen - i));
                }
            }
            return ret.toString();
        }
    }
    

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