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
Squeeze the zigzag pattern horizontally to form a matrix. Now deal with the odd and even columns respectively.
For example let numRows be 5, if we list out the indecies:
row
1 00 08 16
2 01 07 09 15 17
3 02 06 10 14 18
4 03 05 11 13 19
5 04 12 20
First calculate the matrix width:
pairs = floor( len(s) / (numRows + numRows - 2) )
width = pairs * 2 + ceil( (len(s) - pairs * (numRows + numRows - 2)) / numRows )
We can easily make a observation that the direction of odd and even columns and different.
Let the first column be index 0 and let i be the current position at column col.
We need to count the items between matrix[row][col] and matrix[row][col+1], exclusive.
next_i = i + (numRows - row) + (numRows - row), if col is even && 1 < row < numRows
next_i = i + row - 2 + row, if col is odd && 1 < row < numRows
If row == 1 or row == numRows, skip the odd columns.
next_i = i + numRows + (numRows - 2), if col is even && (row == 1 || row == numRows)
/**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
let convert = function (s, numRows) {
if (numRows <= 1) {
return s;
}
const pairs = Math.floor(s.length / (numRows + numRows - 2));
const width =
pairs * 2 +
Math.ceil((s.length - pairs * (numRows + numRows - 2)) / numRows);
let result = "";
for (let row = 1; row <= numRows; row++) {
let i = row - 1;
result += s[i] || "";
for (let col = 0; col < width; col++) {
if (row === 1 || row === numRows) {
if (col % 2 === 0) {
i += numRows + (numRows - 2);
} else {
continue;
}
} else {
if (col % 2 === 0) {
i += numRows - row + (numRows - row);
} else {
i += row - 2 + row;
}
}
result += s[i] || "";
}
}
return result;
};
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☆: .。. o(≧▽≦)o .。.:☆☆: .。. o(≧▽≦)o .。.:☆