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# 6.ZigZag-Conversion

## 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++) {
}
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();
}
}