我的Python解答:
"""
Programmer : EOF
Date : 2015.04.14
File : zigzag.py
E-mail : [email protected]
"""
class Solution:
def convert(self, s, nRows):
length = len(s)
ret_string = ""
i = 0
row = 0
counter = 0
if nRows != 1:
while row < nRows:
i = row
col = 0
while i < length and counter < length:
ret_string += s[i]
if row == 0 or row == nRows -1:
i += (nRows - 1)*2
elif col % 2 == 0:
i += (nRows - row -1)*2
else:
i += row*2
col += 1
counter += 1
row += 1
else:
return s
return ret_string
#----------- just for testing ----------
s = Solution()
print s.convert("AB", 2)
print s.convert("PAYPALISHIRING", 3)
print s.convert("PAYPALISHIRING", 4)凱旋衝鋒的Java解答：
package zigzag_conversion;
public class Solution {
/*
0 [ totalGap ] 6 c
1 [ totalGap-2row ] 5 [ 2row ] 7 b d
2 4 8 a e
3 9 f
*/
public String convert(String s, int nRows) {
if (nRows == 1) {
return s;
}
int len = s.length();
StringBuilder builder = new StringBuilder(len);
int totalGap = (nRows - 1) << 1;
for (int i = 0; i < len; i += totalGap) {
builder.append(s.charAt(i));
}
for (int row = 1; row < nRows - 1; row++) {
for (int i = row, gap = row << 1; i < len; gap = totalGap - gap, i += gap) {
builder.append(s.charAt(i));
}
}
for (int i = nRows - 1; i < len; i += totalGap) {
builder.append(s.charAt(i));
}
return builder.toString();
}
public static void main(String[] args) {
System.out.println(new Solution().convert("ABCD", 4));
}
}皓神的C++解答:
// Source : https://oj.leetcode.com/problems/zigzag-conversion/
// Author : Hao Chen
// Date : 2014-07-17
/**********************************************************************************
*
* 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".
*
*
**********************************************************************************/
#include <iostream>
#include <vector>
#include <string>
using namespace std;
string convert(string s, int nRows) {
//The cases no need to do anything
if (nRows<=1 || nRows>=s.size()) return s;
vector<string> r(nRows);
int row = 0;
int step = 1;
for(int i=0; i<s.size(); i ++) {
if (row == nRows-1) step = -1;
if (row == 0) step = 1;
//cout << row <<endl;
r[row] += s[i];
row += step;
}
string result;
for (int i=0; i<nRows; i++){
result += r[i];
}
return result;
}
int main(int argc, char**argv){
string s;
int r;
s = "PAYPALISHIRING";
r = 3;
cout << s << " : " << convert(s, 3) << endl;
}