leetCode:ZigZag Conversion

最新推荐文章于 2020-03-27 20:46:55 发布

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最新推荐文章于 2020-03-27 20:46:55 发布

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分类专栏：面试 c++ 文章标签：面试 c++ leetCode

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本文介绍了一种将字符串以Z字形方式排列并按行读取的算法实现。该算法接受一个字符串和行数作为输入参数，返回转换后的字符串。通过实例演示了如何使用C++实现这一过程。

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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 text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

#include<iostream>
#include<string>
using namespace std;

class Solution{
public:
    int convert(string s,int nRows);
};

int Solution::convert(string s,int nRows){
    int length=s.length();
    int interval=2*nRows-2;
    for(int i=0;i<nRows;i++){//for
        if(i==0 || i==nRows-1){
        	int m=i;
            int j=0;
            while(interval*j<length){
                cout<<s[interval*j+m];
                j++;
            }
        }else{//else2
            int j=0;int k=j%2;
            int poshead=(nRows-1-i)*2+i;
            int m=0;
            while(i+interval*j/2<length && m*interval+poshead < length ){
                while(i+interval*j/2<length && k==0){
                    cout<<s[i+interval*j/2];
                    j++;k=j%2;
                }
                m=(j-1)/2;
                while(m*interval+poshead < length && k==1){
                    cout<<s[m*interval+poshead];
                    j++;k=j%2;
                }
            }
        }//else2
    }//for
    return 1;
}//convert

int main(){
    string s="0123456789";
    Solution sol;
    sol.convert(s,4);
    return 1;
}