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".
1. Element manipulation.
Special boarder condition on first row and last row
Special boarder condition on 1 row.
Alternative increment, see initial increment to the second step first, will be changed to step 1.
Better way is to setup index,
public class Solution {
public String convert(String s, int numRows) {
int len = s.length();
StringBuilder sb = new StringBuilder();
if(0 == len)
return sb.toString();
if(1 == numRows){
return s;
}
for(int j = 0; j < numRows; ++j){
int d1 = 2*(numRows - 1 - j);
int d2 = 2*j;
if(0 == d1){
for(int i = j; i < len; i += d2){
sb.append(s.charAt(i));
}
}
else if(0 == d2){
for(int i = j; i < len; i += d1){
sb.append(s.charAt(i));
}
}
else{
int d = d2;
for(int i = j; i < len; i += d){
sb.append(s.charAt(i));
if(d == d1)
d = d2;
else
d = d1;
}
}
}
return sb.toString();
}
}
2. The following code works as well, the condition of the last row matches with middle rows, special handling of the first row, first element is OK.
public class Solution {
public String convert(String s, int numRows) {
int len = s.length();
StringBuilder sb = new StringBuilder();
if(0 == len)
return sb.toString();
sb.append(s.charAt(0));
if(1 == numRows){
return s;
}
for(int j = 0; j < numRows; ++j){
int d1 = 2*(numRows - 1 - j);
int d2 = 2*j;
int d = d2;
for(int i = j; i < len; i += d){
if(0 != d){
sb.append(s.charAt(i));
}
if(d == d1)
d = d2;
else
d = d1;
}
}
return sb.toString();
}
}
3. A more efficient code
It turns out, set up a variable to judge the past increment value seems to be the leanest code. Might not be the most efficient one.
It is similar to set a prehead in list manipulation. From somewhere before the list, from somewhere before the array.
public class Solution {
public String convert(String s, int numRows) {
int len = s.length();
StringBuilder sb = new StringBuilder();
if(0 == len)
return sb.toString();
if(1 == numRows){
return s;
}
for(int j = 0; j < numRows; ++j){
int d1 = 2*(numRows - 1 - j);
int d2 = 2*j;
int[] inc = new int[2];
inc[0] = d1;
inc[1] = d2;
int index = 0;
int add = -1;
for(int i = j; i < len; i += inc[index++%2]){
if(0 != add)
sb.append(s.charAt(i));
add = inc[index%2];
}
}
return sb.toString();
}
}
