第3章 栈和队列综合习题(leetcode+vjudge)

用栈实现队列(Implement Stack using Queues)

push(x)

pop()

peek()

empty()

push

pop

empty

top

class MyQueue {
public:
stack<int> st1, st2;
4000
/** Initialize your data structure here. */
MyQueue() {

}

/** Push element x to the back of queue. */
void push(int x) {
st1.push(x);
}

/** Removes the element from in front of queue and returns that element. */
int pop() {
while(!st1.empty()){
st2.push(st1.top());
st1.pop();
}
int result = st2.top();
st2.pop();
while(!st2.empty()){
st1.push(st2.top());
st2.pop();
}
return result;
}

/** Get the front element. */
int peek() {
while(!st1.empty()){
st2.push(st1.top());
st1.pop();
}
int result = st2.top();
while(!st2.empty()){
st1.push(st2.top());
st2.pop();
}
return result;
}

/** Returns whether the queue is empty. */
bool empty() {
return st1.empty();
}
};

用队列实现栈(Implement Stack using Queues )

push(x)

pop()

top()

empty()

push

pop

empty

front

class MyStack {
public:
queue<int> qu1;
queue<int> qu2;
int ptr;
/** Initialize your data structure here. */
MyStack() {

}

/** Push element x onto stack. */
void push(int x) {
qu1.push(x);
}

/** Removes the element on top of the stack and returns that element. */
int pop() {
int counter = 0;
while(!qu1.empty()){
qu2.push(qu1.front());
qu1.pop();
counter++;
}
counter--;        //将计数器自减，以便于取出队尾元素
while(counter > 0){
qu1.push(qu2.front());
qu2.pop();
counter--;
}
int result = qu2.front();
qu2.pop();
return result;
}

/** Get the top element. */
int top() {
int counter = 0;
while(!qu1.empty()){
qu2.push(qu1.front());
qu1.pop();
counter++;
}
counter--;
while(counter > 0){
qu1.push(qu2.front());
qu2.pop();
counter--;
}
int result = qu2.front();
qu2.pop();
qu1.push(result);
return result;
}

/** Returns whether the stack is empty. */
bool empty() {
return qu1.empty();
}
};

最小栈(Min Stack)

min

class MinStack {
public:
/** initialize your data structure here. */
vector<int> _data;
int min;
MinStack() {
min = -1;
}

void push(int x) {
if(_data.empty()){
min = 0;
}
_data.push_back(x);
min = _data.back() < _data[min] ? _data.size() - 1 : min;
}

void pop() {
if(_data.empty()){
return;
}
_data.pop_back();
min = 0;
for(int i = 1; i < _data.size(); ++i){
if(_data[i] < _data[min]){
min = i;
}
}
}

int top() {
return _data.back();
}

int getMin() {
return _data[min];
}
};

字符串解码(Decode String)

k[encoded_string]

encoded_string

k = 1

k

[]

encoded_string

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

class Solution {
public:
string decodeString(string s) {
stack<int> st;
stack<string> st_s;
int num = 0;
int counter;
string dstr = "", tmp;
for(auto it = s.begin(); it != s.end(); ++it){
if(*it >= '0' && *it <= '9'){
num = num * 10 + (*it - '0');
}else if(*it == '['){
st_s.push(dstr);
dstr = "";
st.push(num);
num = 0;
}else if(*it == ']'){
counter = st.top();
st.pop();
tmp = st_s.top();
while(counter--){
tmp += dstr;
}
st_s.pop();
dstr = tmp;
}else{
dstr.push_back(*it);
}
}
return dstr;
}
};

打印机队列(Printer Queue,POJ 3125,NWERC 2006)

第一个任务如果不是当前所有任务中最紧急的，则放到最后；
否则，处理并移除该任务。

Input:
4 2
1 2 3 4
Output:
2

Input:
6 0
1 1 9 1 1 1
Output:
5

#include<cstdio>
#include<algorithm>
#include<deque>
#include<queue>
using namespace std;
struct person
{
int level;
bool me;
person(int lv,int m):level(lv),me(m){}
};
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
priority_queue<int> pq;
deque<person> dq;
int n,mypos,i;
scanf("%d%d",&n,&mypos);
for(i=0;i<n;i++)
{
int tmp;
scanf("%d",&tmp);
if(i==mypos)
dq.push_back(person(tmp,true));
else
dq.push_back(person(tmp,false));
pq.push(tmp);
}
int cnt=1;
while(!pq.empty())
{
if(dq.front().level!=pq.top())
{
dq.push_back(dq.front());
}
else
{
if(dq.front().me) break;
else
{
pq.pop();
cnt++;
}
}
dq.pop_front();
}
printf("%d\n",cnt);
}
return 0;
}

看病要排队(HDU 1873, 2008浙江大学研究生复试全真模拟 )

IN A B

OUT A

OUT A

EMPTY

IN A B

[Input]
7
IN 1 1
IN 1 2
OUT 1
OUT 2
IN 2 1
OUT 2
OUT 1
2
IN 1 1
OUT 1
[Output]
2
EMPTY
3
1
1

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
struct patient
{
int priority;
int time;
patient(int p, int t)
{
priority = p;
time = t;
}
};
struct mycmp
{
bool operator()(patient a, patient b)
{
if (a.priority == b.priority)
return a.time > b.time;
else
return a.priority < b.priority;
}
};
int main()
{
int T;
while (scanf("%d", &T) != EOF)
{
priority_queue<patient, vector<patient>, mycmp> doctors[4];
int id = 0;
while (T--)
{
char ops[10];
scanf("%s", ops);
if (!strcmp(ops, "IN"))
{
int d, p;
scanf("%d %d", &d, &p);
doctors[d].push(patient(p, ++id));
}
else
{
int d;
scanf("%d", &d);
if (doctors[d].empty())
printf("EMPTY\n");
else
{
printf("%d\n", doctors[d].top().time);
doctors[d].pop();
}
}
}
}
return 0;
}

网页浏览器(Web Navigation,POJ 1028)

VISIT 表示浏览某地址
FORWARD 表示前进
BACK 表示后退

#include<iostream>
#include<string>
#include<vector>
#include<stack>
using namespace std;
int main()
{
int cptr = 0, maxsz;
string list[200];
string istr;
list[0] = "http://www.acm.org/";
while (cin >> istr)
{
switch (istr[0])
{
case 'V':
cin >> istr;
list[++cptr] = istr;
maxsz = cptr;
cout << istr << endl;
break;
case 'B':
if (cptr > 0)
{
cout << list[--cptr] << endl;
}
else
{
cptr = 0;
cout << "Ignored" << endl;
}
break;
case 'F':
if (cptr == maxsz)
{
cout << "Ignored" << endl;
}
else
{
cout << list[++cptr] << endl;
}
break;
case 'Q':
return 0;
}
}
return 0;
}