[LeetCode] 6. ZigZag Converesion 之字型转换字符串

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".

这道题就是看坐标的变化，找规律并分块处理。参考：爱做饭的小莹子

规律：第一行和最后一行，就是按照2n-2的顺序一点点加的。斜着那条线的字的位置是当前列j+(2n-2)-2i(i是行的index）。

Java：

public String convert(String s, int nRows) {  
        if(s == null || s.length()==0 || nRows <=0)  
            return "";  
        if(nRows == 1)  
            return s;
            
        StringBuilder res = new StringBuilder();  
        int size = 2*nRows-2;  
        for(int i=0;i<nRows;i++){  
            for(int j=i;j<s.length();j+=size){  
                res.append(s.charAt(j));  
                if(i != 0 && i != nRows - 1){//except the first row and the last row
                    int temp = j+size-2*i;
                    if(temp<s.length())
                        res.append(s.charAt(temp));
                }
            }                  
        }  
        return res.toString();  
    }

Python:

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        if numRows == 1:
            return s
        step, zigzag = 2 * numRows - 2, ""
        for i in xrange(numRows):
            for j in xrange(i, len(s), step):
                zigzag += s[j]
                if 0 < i < numRows - 1 and j + step - 2 * i < len(s):
                    zigzag += s[j + step - 2 * i]
        return zigzag

C++:

class Solution {
public:
    string convert(string s, int nRows) {
        if (nRows <= 1) return s;
        string res = "";
        int size = 2 * nRows - 2;
        for (int i = 0; i < nRows; ++i) {
            for (int j = i; j < s.size(); j += size) {
                res += s[j];
                int tmp = j + size - 2 * i;
                if (i != 0 && i != nRows - 1 && tmp < s.size()) res += s[tmp];
            }
        }
        return res;
    }
};

　　

　　

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