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".
思路:
n=2时,字符串坐标变成zigzag的走法就是:
0 2 4 6
1 3 5 7
n=3时的走法是:
0 4 8
1 3 5 7 9
2 6 10
n=4时的走法是:
0 6 12
1 5 7 11 13
2 4 8 10 14
3 9 15
利用这个规律,可以按行填字,第一行和最后一行,就是按照2n-2的顺序排列。
其他行除了上面那个填字规则,就是还要处理斜着那条线的字,可以发现那条线的字的位置永远是当前列j+(2n-2)-2i(i是行的index),同时和 2n-2间隔排列组成中间的行
代码:
public class Solution {
public String convert(String s, int numRows) {
if(s == null && s.length() == 0 && numRows <= 0){
return "";
}
if(numRows == 1){
return s;
}
StringBuffer sb = new StringBuffer();
int size = 2*numRows - 2;
for(int i = 0;i < numRows;i++){
for(int j = i;j < s.length();j = j+size){
sb.append(s.charAt(j));
if(i != 0 && i != numRows-1){
int tmp = j + size -2*i;
if(tmp < s.length()){
sb.append(s.charAt(tmp));
}
}
}
}
return sb.toString();
}
}