6. ZigZag Conversion
Description:
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"
.
Solutions:
If we read the characters in zigzag way, we will find the row numbers periodically
repeat. For rownumber > 1
, the period is (rownumber-1)*2
. For example, if rownumber = 6
, s = "PAYPALISHIRINGAROUNDTHEWORLD"
, the pattern is
0 P R H
1 A I I T E
2 Y H N D W
3 P S G N O
4 A I A O R D
5 L R L
The row numbers repeat as "0 1 2 3 4 5 4 3 2 1"
, with length equal to 10 (= 5*2)
. It's not hard for us to find the repeated pattern, which starts from 0
to n-1
and then decreases to 1
.
So, we can use a StringBuffer array sb
to represent each line's string and read through the string and append the character to sb[0], sb[1], ..., sb[n-1], sb[n-2], ... sb[1]
periodically. Meanwhile, an array is used to indicate which StringBuffer we shoud append to.
Here is the solution:
public class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) return s;
StringBuffer[] sb = new StringBuffer[numRows];
// init sb
for (int i = 0; i < numRows; i++) {
sb[i] = new StringBuffer();
}
int period = (numRows-1)*2;
int[] index = new int[period];
// init index
for (int i = 0; i <= period/2; i++) {
index[i] = i;
}
for (int i = period-1, j = 1; i > period/2; i--) {
index[i] = j++;
}
for (int i = 0; i < s.length(); i++) {
sb[index[i%period]].append(s.charAt(i));
}
StringBuffer res = new StringBuffer();
for (int i = 0; i < numRows; i++) {
res = res.append(sb[i]);
}
return res.toString();
}
}