给定两个队列A和B ，每个队列的大小为N ，任务是根据B中的执行顺序找到执行A 中任务所需的最短时间，其中：
- 如果在队列B前面找到的任务在队列A前面，则弹出该任务并执行它。
- 如果在队列B前面找到的任务在队列A前面没有找到，则从队列A 中弹出当前任务并将其推送到末尾。
- 队列中的 push 和 pop 操作花费一个单位时间，任务的执行是在恒定时间内完成的。
例子
Input: A = { 3, 2, 1 }, B = { 1, 3, 2 }
Output: 7
Explanation:
For A = 3 and B = 1 => Since the front of queue A does not match with the front of queue B. Pop and Push 3 at the back of Queue. Then, A becomes { 2, 1, 3 } and time consumed = 2 ( 1 unit of time for each push and pop operation)
For A = 2 and B = 1 => Since the front of queue A does not match with the front of queue B. Pop and Push 2 at the back of Queue. Then, A becomes { 1, 3, 2 } and time = 2 + 2 = 4
For A = 1 and B = 1 => Since the front of queue, A equals to front of queue B. Pop 1 from both the queue and execute it, Then A becomes { 3, 2 } and B becomes { 3, 2 } and time = 4 + 1 = 5
For A = 3 and B = 3 => Since the front of queue, A equals to front of queue B. Pop 3 from both the queue and execute it, Then A becomes { 2 } and B becomes { 2 } and time = 5 + 1 = 6
For A = 2 and B = 2 => Since the front of the queue, A equals to front of queue B. Pop 2 from both the queue and execute it. All the tasks are executed. time = 6 + 1 = 7
Therefore the total time is 7.
Input: A = { 3, 2, 1, 4 }, B = { 4, 1, 3, 2 }
Output: 14
方法：
对于队列A 中的每个任务：
- 如果队列 A 的前端任务与队列B的前端任务相同。从两个队列中弹出任务并执行它。将总时间增加一个单位。
- 如果队列 A 的前端任务与队列B的前端任务不同。从队列 A 中弹出任务并将其推送到队列 A 的后面，并将总时间增加两个单位。 (1 个用于弹出操作，1 个用于推送操作)
- 重复以上步骤，直到队列A中的任务全部执行完毕。
下面是上述方法的实现：
CPP
// C++ program to find the total
// time taken to execute the task
// in given order
#include "bits/stdc++.h"
using namespace std;
// Function to calculate the
// total time taken to execute
// the given task in original order
int run_tasks(queue& A,
queue& B)
{
// To find the total time
// taken for executing
// the task
int total_time = 0;
// While A is not empty
while (!A.empty()) {
// Store the front element of queue A and B
int x = A.front();
int y = B.front();
// If the front element of the queue A
// is equal to the front element of queue B
// then pop the element from both
// the queues and execute the task
// Increment total_time by 1
if (x == y) {
A.pop();
B.pop();
total_time++;
}
// If front element of queue A is not equal
// to front element of queue B then
// pop the element from queue A &
// push it at the back of queue A
// Increment the total_time by 2
//(1 for push operation and
// 1 for pop operation)
else {
A.pop();
A.push(x);
total_time += 2;
}
}
// Return the total time taken
return total_time;
}
// Driver Code
int main()
{
// Given task to be executed
queue A;
A.push(3);
A.push(2);
A.push(1);
A.push(4);
// Order in which task need to be
// executed
queue B;
B.push(4);
B.push(1);
B.push(3);
B.push(2);
// Function the returns the total
// time taken to execute all the task
cout << run_tasks(A, B);
return 0;
} Java
// Java program to find the total
// time taken to execute the task
// in given order
import java.util.*;
class GFG
{
// Function to calculate the
// total time taken to execute
// the given task in original order
static int run_tasks(Queue A,
Queue B)
{
// To find the total time
// taken for executing
// the task
int total_time = 0;
// While A is not empty
while (!A.isEmpty())
{
// Store the front element of queue A and B
int x = A.peek();
int y = B.peek();
// If the front element of the queue A
// is equal to the front element of queue B
// then pop the element from both
// the queues and execute the task
// Increment total_time by 1
if (x == y)
{
A.remove();
B.remove();
total_time++;
}
// If front element of queue A is not equal
// to front element of queue B then
// pop the element from queue A &
// push it at the back of queue A
// Increment the total_time by 2
//(1 for push operation and
// 1 for pop operation)
else
{
A.remove();
A.add(x);
total_time += 2;
}
}
// Return the total time taken
return total_time;
}
// Driver Code
public static void main(String[] args)
{
// Given task to be executed
Queue A = new LinkedList();
A.add(3);
A.add(2);
A.add(1);
A.add(4);
// Order in which task need to be
// executed
Queue B = new LinkedList();
B.add(4);
B.add(1);
B.add(3);
B.add(2);
// Function the returns the total
// time taken to execute all the task
System.out.print(run_tasks(A, B));
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 program to find the total
# time taken to execute the task
# in given order
from collections import deque
# Function to calculate the
# total time taken to execute
# the given task in original order
def run_tasks(A, B):
# To find the total time
# taken for executing
# the task
total_time = 0
# While A is not empty
while (len(A) > 0):
# Store the front element of queue A and B
x = A.popleft()
y = B.popleft()
# If the front element of the queue A
# is equal to the front element of queue B
# then pop the element from both
# the queues and execute the task
# Increment total_time by 1
if (x == y):
total_time += 1
# If front element of queue A is not equal
# to front element of queue B then
# pop the element from queue A &
# append it at the back of queue A
# Increment the total_time by 2
#(1 for append operation and
# 1 for pop operation)
else:
B.appendleft(y)
A.append(x)
total_time += 2
# Return the total time taken
return total_time
# Driver Code
if __name__ == '__main__':
# Given task to be executed
A = deque()
A.append(3)
A.append(2)
A.append(1)
A.append(4)
# Order in which task need to be
# executed
B = deque()
B.append(4)
B.append(1)
B.append(3)
B.append(2)
# Function the returns the total
# time taken to execute all the task
print(run_tasks(A, B))
# This code is contributed by mohit kumar 29C#
// C# program to find the total
// time taken to execute the task
// in given order
using System;
using System.Collections.Generic;
class GFG
{
// Function to calculate the
// total time taken to execute
// the given task in original order
static int run_tasks(Queue A,
Queue B)
{
// To find the total time
// taken for executing
// the task
int total_time = 0;
// While A is not empty
while (A.Count != 0)
{
// Store the front element of queue A and B
int x = A.Peek();
int y = B.Peek();
// If the front element of the queue A
// is equal to the front element of queue B
// then pop the element from both
// the queues and execute the task
// Increment total_time by 1
if (x == y)
{
A.Dequeue();
B.Dequeue();
total_time++;
}
// If front element of queue A is not equal
// to front element of queue B then
// pop the element from queue A &
// push it at the back of queue A
// Increment the total_time by 2
//(1 for push operation and
// 1 for pop operation)
else
{
A.Dequeue();
A.Enqueue(x);
total_time += 2;
}
}
// Return the total time taken
return total_time;
}
// Driver Code
public static void Main(String[] args)
{
// Given task to be executed
Queue A = new Queue();
A.Enqueue(3);
A.Enqueue(2);
A.Enqueue(1);
A.Enqueue(4);
// Order in which task need to be
// executed
Queue B = new Queue();
B.Enqueue(4);
B.Enqueue(1);
B.Enqueue(3);
B.Enqueue(2);
// Function the returns the total
// time taken to execute all the task
Console.Write(run_tasks(A, B));
}
}
// This code is contributed by PrinciRaj1992 输出：
14
时间复杂度： O(N 2 )，其中 N 是任务数。
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