n个数，入栈出栈，所有情况

Posted 2022-11-26 nirvana · rebirth

int count = 0;
void print(vector<int>& trace, vector<int>& push_order) 
	int i = 0;
	stack<int>stk;
	for (auto n : trace) 
		if (n == 1) 
			stk.push(push_order[i++]);
		
		else 
			int num = stk.top();
			stk.pop();
			cout << num << " ";
		
	
	cout << endl;
	::count++;


void backtrack(vector<int>& trace, vector<int>& push_order, int left)  // trace:0 出栈,1 入栈, push_order:操作的数字, left:push比pop多的次数，也就是栈中还有几个数
	//    left记录前缀序列中1的个数，每当trace中push一个0，就减一
	if (trace.size() == 2 * push_order.size())  // 递归出口
		if (left == 0) // 只有left=0，才能使得0的数量==1的数量，才是合法的。
			print(trace, push_order);
		return;
	
	// 做选择
	if (left > 0)  // 如果left>0，则可以push 0进去 ,同时 left-1
		trace.push_back(0);
		backtrack(trace, push_order, left - 1);
		// 撤销选择
		trace.pop_back();
	
	trace.push_back(1);
	backtrack(trace, push_order, left + 1);
	trace.pop_back();

int main() 
	vector<int> push_order =  1,2,3,4 ;
	vector<int>trace;
	backtrack(trace, push_order, 0);
	cout << ::count << endl;

	const int maxI = 4;
	function<void(stack<int>&, list<int>&, int)> fun = [&](stack<int> st, list<int> out, int index) 
		if (out.size() == maxI) 
			for_each(begin(out), end(out), [](int x) cout << x << " "; ), cout << endl;
			return;
		
		if (st.size() > 0) 
			int t = st.top();	st.pop();	out.push_back(t);
			fun(st, out, index);
			st.push(t);	out.pop_back();
		
		if (index <= maxI) 
			st.push(index++);
			fun(st, out, index);
		
	;
	stack<int>st;
	list<int> out;
	fun(st, out, 1);