6. Zigzag Conversion

Medium (Trung bình) C++ Python 🔗 Xem trên LeetCode

📋 Đề Bài

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

Example 3:

Input: s = "A", numRows = 1
Output: "A"

 

Constraints:

  • 1 <= s.length <= 1000
  • s consists of English letters (lower-case and upper-case), ',' and '.'.
  • 1 <= numRows <= 1000

🧠 Thuật Toán & Kỹ Thuật

String (Chuỗi)
⏱️ Thời gian O(n²)
💾 Không gian O(n)

💻 Lời Giải

C++ 0006-zigzag-conversion.cpp 
class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) {
            return s;
        }
        vector<string> listStr(numRows);
        int n = (int)s.size();
        int step = 1;
        int k = 0;
        for (int i = 0; i < n; ++i) {
            listStr[k] += s[i];
            if (k == numRows - 1) {
                step = -1;
            }
            else if (k == 0) {
                step = 1;
            }
            k += step;
        }
        string res;
        for (string str : listStr) {
            res += str;
        }
        return res;
    }
};
Python 0006-zigzag-conversion.py 
class Solution:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1:
            return s
        
        ans = [''] * numRows 
        idx, step = 0, 1
        
        for c in s:
            ans[idx] += c
            if idx == 0:
                step = 1
            elif idx == numRows - 1:
                step = -1
            idx += step
            
        return ''.join(ans)