6. ZigZag Conversion

6. ZigZag Conversion

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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".　　

 

又跟上次pat的书法字符串很相似，划成Z型的字符串。找出规律后正向读进去就行了。

 

 

首先第一行和最后一个行的间隔是2*n-2个元素

中间行，分布规律间隔interval - 2*i，2*i，interval - 2*i，2*i，interval - 2*i 。。。。。（交替）<len个元素

 注意i是从0开始的，从1开始就减去1.

 

#include <bits/stdc++.h>

#define INF 0x3fffffff
#define eps 1e-8

typedef long long LL;
const double pi = acos(-1.0);
const int mod = 1e9 + 7;
const int maxn = 70;
using namespace std;

string convert(string s, int nRows)
{
    if(nRows == 1)return s;
    int len = s.size(), k = 0, interval = (nRows<<1)-2;
    string res(len, ' ');
    for(int j = 0; j < len ; j += interval)//处理第一行
        res[k++] = s[j];
    for(int i = 1; i < nRows-1; i++)//处理中间行
    {
        int inter = (i<<1);//就是i*2，写成位运算是不是高大上一些
        for(int j = i; j < len; j += inter)
        {
            //第一次加的就是字符串第一列的
            res[k++] = s[j];
            inter = interval - inter;//interval–2*i或者interval-(interval–2*i)=2*i
        }
    }
    for(int j = nRows-1; j < len ; j += interval)//处理最后一行
        res[k++] = s[j];
    return res;
}

int main()
{
    freopen("in.txt","r",stdin);
    string ss;
    cin>>ss;
    cout<<convert(ss, 5);
    return 0;
}