蜗牛慢慢爬 LeetCode 6. ZigZag Conversion [Difficulty: Medium]

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);

翻译

此题坑爹在于关于zigzag的形式没有说清楚 然后在leetcode上负分较多

其实只要理解了zigzag的形式就好了

/*n=numRows
Δ=2n-2    1                           2n-1                         4n-3
Δ=        2                     2n-2  2n                    4n-4   4n-2
Δ=        3               2n-3        2n+1              4n-5       .
Δ=        .           .               .               .            .
Δ=        .       n+2                 .           3n               .
Δ=        n-1 n+1                     3n-3    3n-1                 5n-5
Δ=2n-2    n                           3n-2                         5n-4
*/

Hints

Related Topics: String

一开始我的想法是用二维数组在遍历的时候把字符存到对应的位置 之后再join到一个字符串里面 这样挺直观的 但是python的话慢了一点

事实上我的代码有点硬推出每个字符的位置填上去的意思(囧) discuss中类似状态机改变index变化步长的方法显然更好

代码

Java

public String convert(String s, int nRows) {
    char[] c = s.toCharArray();
    int len = c.length;
    StringBuffer[] ss = new StringBuffer[nRows];
    for (int i = 0; i < ss.length; i++) ss[i] = new StringBuffer();

    int i = 0;
    while (i < len) {
        for (int idx = 0; idx < nRows && i < len; idx++)
            ss[idx].append(c[i++]);
        for (int idx = nRows-2; idx >= 1 && i < len; idx--)
            ss[idx].append(c[i++]);
    }
    for (int idx = 1; idx < ss.length; idx++)
        ss[0].append(ss[idx]);
    return ss[0].toString();
}

Python

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        l = len(s)
        if l<=1 or numRows<=1:   return s
        k = numRows - 2
        n = l/(numRows+k)+1
        n = n*(k+1)

        zigzag = [['']*n for i in range(numRows)]
        for i in range(l):
            c = (2*numRows-2)*(i/(2*numRows-2))
            if (i-c)/numRows==0:
                zigzag[((i-c)%numRows)][(i/(2*numRows-2))*(k+1)] = s[i]
            else:
                zigzag[numRows-2-(i-c)%numRows][(i/(2*numRows-2))*(k+1)+((i-c)%numRows+1)] = s[i]
        ss =''
        for i in range(numRows):
            ss += ''.join(zigzag[i])

        return ss

#better solution
class Solution(object):
    def convert(self, s, numRows):
        if numRows==1or numRows>=len(s):
            return s
        L = ['']*numRows
        index, step = 0, 1

        for i in s:
            L[index] += i
            if index==0:
                step = 1
            elif index==numRows-1:
                step = -1
            index += step
        return ''.join(L)