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 RAnd 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
The String ZigZag Conversion Algorithms
We can simulate the process by walking down first, then when it reaches the last row, bounces back in the reverse direction. We need arrays of strings as we walk, and append the characters to the corresponding row of string.
O(N) time and O(N) space requirement. Edge case is when the ZigZag row is one and then we have to return the original string.
C++ String ZigZag Conversion
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | class Solution { public: string convert(string s, int numRows) { if (numRows == 1) return s; vector<string> rows(min((int)s.size(), numRows)); bool down = false; int row = 0; for (int i = 0; i < s.size(); ++ i) { rows[row] += s[i]; if ((row == 0) || (row == numRows - 1)) { down = !down; } row += down ? 1 : -1; } string ss = ""; for (const auto &n: rows) { ss += n; } return ss; } }; |
class Solution {
public:
string convert(string s, int numRows) {
if (numRows == 1) return s;
vector<string> rows(min((int)s.size(), numRows));
bool down = false;
int row = 0;
for (int i = 0; i < s.size(); ++ i) {
rows[row] += s[i];
if ((row == 0) || (row == numRows - 1)) {
down = !down;
}
row += down ? 1 : -1;
}
string ss = "";
for (const auto &n: rows) {
ss += n;
}
return ss;
}
};Python String ZigZag Conversion
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | class Solution: def convert(self, s: str, numRows: int) -> str: if numRows == 1: return s data = [''] * numRows down = False row = 0 for i in s: data[row] += i if row == 0 or row == numRows - 1: down = not down row = row + 1 if down else row - 1 x = '' for j in data: if j != '': x = x + j return x |
class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
data = [''] * numRows
down = False
row = 0
for i in s:
data[row] += i
if row == 0 or row == numRows - 1:
down = not down
row = row + 1 if down else row - 1
x = ''
for j in data:
if j != '':
x = x + j
return xJavascript String ZigZag Conversion
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | /** * @param {string} s * @param {number} numRows * @return {string} */ var convert = function(s, numRows) { if (numRows === 1) return s; let data = []; let row = 0; let down = false; for (let x of s) { if (typeof data[row] === 'undefined') { data[row] = x; } else { data[row] += x; } if (row === 0 || row === numRows - 1) { down = !down; } row += down ? 1 : -1; } let res = ''; for (let i = 0; i < Math.min(numRows, s.length); ++ i) { res += data[i]; } return res; }; |
/**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
var convert = function(s, numRows) {
if (numRows === 1) return s;
let data = [];
let row = 0;
let down = false;
for (let x of s) {
if (typeof data[row] === 'undefined') {
data[row] = x;
} else {
data[row] += x;
}
if (row === 0 || row === numRows - 1) {
down = !down;
}
row += down ? 1 : -1;
}
let res = '';
for (let i = 0; i < Math.min(numRows, s.length); ++ i) {
res += data[i];
}
return res;
};–EOF (The Ultimate Computing & Technology Blog) —
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