Comment on page

6.ZigZag-Conversion

6. ZigZag Conversion

题目地址

题目描述

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 s, int numRows);
​
Example 1:
Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
​
Example 2:
Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
​
P I N
A L S I G
Y A H R
P I

代码

Approach #1 Sort by Row

class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) return s;
​
List<StringBuilder> rows = new ArrayList();
for (int i = 0; i < Math.min(numRows, s.length()); i++) {
rows.add(new StringBuilder());
}
​
int curRow = 0;
boolean goingDown = false;
​
for (char c: s.toCharArray()) {
rows.get(curRow).append(c);
if (curRow == 0 || curRow == numRows - 1) {
goingDown = !goingDown;
}
curRow += goingDown ? 1 : -1;
}
​
StringBuilder ret = new StringBuilder();
for (StringBuilder row : rows) {
ret.append(row);
}
return ret.toString();
}
}

Approach #2 Visit By Row

class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) return s;
StringBuilder ret = new StringBuilder();
int n = s.length();
int cycleLen = 2 * numRows - 2;
​
for (int i = 0; i < numRows; i++) {
for (int j = 0; j + i < n; j += cycleLen) {
ret.append(s.charAt(j + i));
// inner row
if (i != 0 && i != numRows - 1 && j + cycleLen - i < n) {
ret.append(s.charAt(j + cycleLen - i));
}
}
}
​
return ret.toString();
}
}