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：通过观察这个N型的图案的组成，我们发现可以通过从上到下（0 ~ numRows - 1），再从下到上(numRows - 1, 1)，以此为反复，就可以得到想要的排列。所以解决的方法就是创建numRows个StringBuilder，每个StringBuilder都可以储存每行的字母，通过N型的排列来将字母一个个加进对应的行里。
public String convert(String
s,
int numRows) {
if (
s.
length() ==
0 || numRows <=
1)
return s;
int length =
s.
length(),
index =
0;
StringBuilder[] sb = new StringBuilder[numRows];
for (
int i =
0; i < numRows; i++) {
sb[i] = new StringBuilder();
}
while (
index <
length) {
for (
int i =
0; i < numRows &&
index <
length; i++) {
System.out.println(sb[i]);
sb[i].append(
s.charAt(
index));
index++;
}
for (
int j = numRows -
2; j >=
1 &&
index <
length; j--) {
sb[j].append(
s.charAt(
index));
index++;
}
}
StringBuilder res = new StringBuilder();
for (
int i =
0; i < numRows; i++) {
res.append(sb[i]);
}
return res.toString();
}
