LeetCode Algorithm 0006 - ZigZag Conversion (Medium)

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本文详细介绍了如何实现LeetCode上的ZigZag Conversion算法题，该题要求将输入字符串以Z字形方式分布于指定行数上，再按行读取形成新的字符串。通过C++代码示例，讲解了算法的实现过程，包括错误处理和特殊情况的考虑。

LeetCode Algorithm 0006 - ZigZag Conversion (Medium)

Problem Link: https://leetcode.com/problems/zigzag-conversion/description/

文章目录

LeetCode Algorithm 0006 - ZigZag Conversion (Medium)
- Description
- Solution C++

Description

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

Solution C++

#pragma once

#include "pch.h"

// Problem: https://leetcode.com/problems/zigzag-conversion/description/

namespace P6ZigZagConversion
{
    class Solution
    {
        public:
        string convert(string s, int numRows)
        {
            if (numRows < 1)
            {
                logic_error e = logic_error("`numRows` must be greater than zero.");
                throw e;
            }

            if (numRows == 1)
            {
                return s;
            }

            vector<string> lines = vector<string>(numRows, "");
            bool add = true;
            for (size_t i = 0, li = 0; i < s.size(); i++)
            {
                lines[li].push_back(s[i]);
                li = add ? li + 1 : li - 1;
                
                if (li == numRows || li == -1)
                {
                    li = add ? numRows - 2 : 1;
                    add = !add;
                }
            }

            string result = "";
            for (vector<string>::iterator it = lines.begin(); it != lines.end(); it++)
            {
                result += *it;
            }

            return result;
        }
    };
}