LeetCode--ZigZag Conversion

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ZigZag Conversion

题目

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

分析

这题涉及到ZigZag这个知识点，从题目我们不得而知这是个什么东西，从wiki上，我们可以了解到它大概是这个东西

再看看解释

A zigzag is a pattern made up of small corners at variable angles, though constant within the zigzag, tracing a path between two parallel lines; it can be described as both jagged and fairly regular.

From the point of view of symmetry, a regular zigzag can be generated from a simple motif like a line segment by repeated application of a glide reflection.

The origin of the word is unclear. Its first printed appearance was in French books in the late 17th century.[1]

用我们自己的话讲，就是一个锯齿形的数据模式，本题的要求是将锯齿形的数据（给定行数）复原为原来的数据。这题着重考察的是数学关系的发现，我们采取的策略是从锯齿的最下面那个点开始遍历，给出得函数也是数学关系的一部分，这样我们发现的数学关系如下：

   1. 最下面数据的下标是(numRows-1)+[numRows+(numRows-2)]*N.

   2. 根据这个数学关系和numRows，我们可以得出每一行字符的排列，就是通过最下面的下标index+i或者index-i得出。

源码

class Solution {
public:
    string convert(string s, int numRows) {
        string result;
        int m = 0;
        int math = numRows + (numRows - 2);
        vector<vector<char> > v(numRows, vector<char>());
        if(numRows == 1||s.size() < numRows) {
        	return s;
        }
        for(int i = (numRows-1); i < s.size(); i = i+math) {
        	for(int j = 0; j < numRows; j++) {
        		if(j == 0) {
        			v[j].push_back(s[i]);
        		} else {
        			int l = i - j;
        			int r = i + j;
        			v[j].push_back(s[l]);
        			if(j != numRows - 1 &&r < s.size())
        				v[j].push_back(s[r]);
        		}
        	}
        	m = i;
        }

        if(m+numRows-1 < s.size()) {
        	for(int i = 0; i < numRows; i++) {
        		if(m+numRows-1+i < s.size()) {
        			v[numRows-1-i].push_back(s[m+numRows-1+i]);
        		}
        	}	
        }
        for(int i = numRows-1; i >= 0; i--) {
        	for(int j = 0; j < v[i].size(); j++) {
        		result.push_back(v[i][j]);
        	}
        }
        return result;
    }
};