LEECODE Algorithm 6. ZigZag Conversion

• 题目
• 代码块
• 想法

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)

Find the median of the two sorted arrays. The overall run time complexity >should be O(log (m+n)).

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 text, int nRows);

convert(“PAYPALISHIRING”, 3) should return “PAHNAPLSIIGYIR”.

代码

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows == 1){
            return s;
        }
        int quotient = s.length()/(2*numRows-2);
        int remainder = s.length()%(2*numRows-2);
        int numCols = 0;
        if(remainder > numRows){
            numCols = 1 + remainder - numRows + (numRows - 1)*quotient;
        }
        else{
            numCols = 1 + (numRows - 1)*quotient;
        }
        char arr[numRows][numCols];
        string output = "";
        int colcount = 0,down = 1,k = 0;;
        for(int i =0; i < numRows; i++){
            for(int j = 0; j < numCols; j++){
                arr[i][j] = '0';
            }
        }
        for(int i = 0; i < s.length(); i++){
            k++;
            if(down == 1){
                arr[k-1][colcount] = s[i];
            }
            else{
                arr[numRows-k-1][colcount] = s[i];
            }

            if(down == 0){
                colcount++;
            }
            else{
                if(k == numRows){
                    colcount++;
                    down = 0;
                    k = 0;
                }
            }
            if(down == 0 && k == numRows-2){
                k = 0;
                down = 1;
            }

        }
        for(int i =0; i < numRows; i++){
            for(int j = 0; j < numCols; j++){
                if(arr[i][j] != '0'){
                    output += arr[i][j];
                }
            }
        }       
        return output;
    }
};

想法

首先，很明显这一题用矩阵来记录z字型信息，关键是求得矩阵的row，col值。对于n行的z字形，完全的上升部分有n-2个点，占用了n-2列，通过数学知识易得row，col。随后设置一个s.length()次的for循环，设置变量colcount，记录下一次字符应该放的位置，依次，便可得出结果.