[LeetCode]6. ZigZag ConversionZ

2018-12-26  本文已影响0人  jchen104

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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