6. ZigZag Conversion
Problem's Link
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Mean:
给你一个字符串,让你将其按照倒‘之’字型排列,然后输出排列后的顺序.
analyse:
简单的推公式,算出随行递增,间隔的变化.(第一行和最后一行特判一下)
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-02-15-15.00
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
string convert(string s, int nRows)
{
if(nRows <= 1 || s.length() < 3 || s.length() <= nRows) return s;
string s2;
int zigSpan = nRows*2 - 2;
for (int i = 0; i < nRows; i++)
{
for (int j = i; j < s.length(); j+=zigSpan)
{
s2.push_back(s[j]);
if (i != 0 && i != nRows-1 && zigSpan+j-2*i<s.length())
s2.push_back(s[zigSpan+j-2*i]);
}
}
return s2;
}
};
int main()
{
return 0;
}
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-02-15-15.00
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
string convert(string s, int nRows)
{
if(nRows <= 1 || s.length() < 3 || s.length() <= nRows) return s;
string s2;
int zigSpan = nRows*2 - 2;
for (int i = 0; i < nRows; i++)
{
for (int j = i; j < s.length(); j+=zigSpan)
{
s2.push_back(s[j]);
if (i != 0 && i != nRows-1 && zigSpan+j-2*i<s.length())
s2.push_back(s[zigSpan+j-2*i]);
}
}
return s2;
}
};
int main()
{
return 0;
}