6. ZigZag Conversion

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".

链接: http://leetcode.com/problems/zigzag-conversion/

题解: 

根据例子列出n = 3和n = 4的情况，计算出列间距zigSize = 2 * numRows - 2，所以每行元素为 i + n * zigSize以及斜线的计算公式，当i为行数时，斜线上元素为i + n * zigSize - 2 * i。

Time Complexity - O(n)， Space Complexity - O(n)。

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

 

二刷:

Java:

要注意 zigSize = 2 * numSize - 2,  zag = j + zigSize - 2 * i

public class Solution {
    public String convert(String s, int numRows) {
        int zigSize = 2 * numRows - 2;  // calculate zigSize
        if (s == null || s.length() == 0 || zigSize <= 0) {
            return s;
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < numRows; i++) {
            for (int j = i; j < s.length(); j += zigSize) {
                sb.append(s.charAt(j));
                if (i != 0 && i != numRows - 1 && j + zigSize - 2 * i < s.length()) { // calculate zag
                    sb.append(s.charAt(j + zigSize - 2 * i));
                }
            }
        }    
        return sb.toString();   
    }
}

 

Python:

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        zigSize = 2 * numRows - 2
        if s == None or zigSize <= 0:
            return s
        res = ''
        for i in range(numRows):
            for j in range(i, len(s), zigSize):
                res += s[j]
                zag = j + zigSize - 2 * i
                if i not in (0, numRows - 1) and zag < len(s):
                    res += s[zag]
        return res
                    

 

三刷:

其实就是看了一遍二刷的解而已。

Java:

public class Solution {
    public String convert(String s, int numRows) {
        int zigSize = 2 * numRows - 2; 
        if (s == null || s.length() == 0 || zigSize <= 0) return s;
        
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < numRows; i++) {
            for (int j = i; j < s.length(); j += zigSize) {
                sb.append(s.charAt(j));
                if (i != 0 && i != numRows - 1 && j + zigSize - 2 * i < s.length()) { // calculate zag
                    sb.append(s.charAt(j + zigSize - 2 * i));
                }
            }
        }    
        return sb.toString();   
    }
}

 

Reference:

http://www.cnblogs.com/springfor/p/3889414.html