107 lines
2.1 KiB
C++
107 lines
2.1 KiB
C++
/*
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VIM: let b:lcppflags="-std=c++14 -O2 -pthread -I."
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VIM: let b:wcppflags="/O2 /EHsc /DWIN32"
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VIM-: let b:cppflags=g:Iboost.g:Itbb
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VIM-: let b:ldflags=g:Lboost.g:Ltbb
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VIM-: let b:ldlibpath=g:Bboost.g:Btbb
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VIM-: let b:argv=""
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*/
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#include <iostream>
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#include <exception>
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#include <chrono>
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#include <vector>
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#include <algorithm>
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/*Given N and M find all stepping numbers in range N to M
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The stepping number:
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A number is called as a stepping number if the adjacent digits have a difference of 1.
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e.g 123 or 101 is stepping number, but 358 is not a stepping number
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Example:
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N = 10, M = 20
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all stepping numbers are 10 , 12
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Return the numbers in sorted order.
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*/
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/*
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* Solution: complexity is linear actually. O(n)
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* So, a better solution I guess will be to interate over the range
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* and check each number on condition.
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*/
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void f( std::vector<int>& v, int A, int B, int n )
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{
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if ( A <= n && n <= B ) {
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v.push_back(n);
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}
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else if ( n > B ) {
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return;
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}
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int d = n % 10;
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if ( d + 1 <= 9 ){
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f( v, A, B, n*10+d+1);
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}
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if ( d - 1 >= 0 ){
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f( v, A, B, n*10+d-1);
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}
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}
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std::vector<int> stepnum(int A, int B)
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{
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std::vector<int> v;
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if ( A <= 0 && 0 <= B ){
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v.push_back(0);
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}
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for ( int i = 1; i <= 9; ++i ){
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f( v, A, B, i );
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}
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std::sort(v.begin(), v.end());
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return v;
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}
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void print( const std::vector<int>& A )
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{
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for (auto i: A){
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std::cout << i << " ";
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}
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std::cout << std::endl;
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}
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int main ( void )
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{try{
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auto begin = std::chrono::high_resolution_clock::now();
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//......
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print( stepnum(10, 20) );
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print( stepnum(0, 1000) );
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auto end = std::chrono::high_resolution_clock::now();
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std::chrono::duration<double> seconds = end - begin;
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std::cout << "Time: " << seconds.count() << std::endl;
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return 0;
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}
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catch ( const std::exception& e )
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{
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std::cerr << std::endl
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<< "std::exception(\"" << e.what() << "\")." << std::endl;
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return 2;
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}
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catch ( ... )
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{
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std::cerr << std::endl
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<< "unknown exception." << std::endl;
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return 1;
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}}
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