Backtracking-how to solve subset problem and full permutation

本文介绍了两种常见的算法——全排列问题与子集问题,并提供了详细的实现案例。针对全排列问题,采用邻位交换法与标签法两种解决方案;对于子集问题,则运用选择拼接法实现。

在算法基础中,有两种关于排列的算法经常使用到,一种是<1>全排列问题(有序),一种是<2>子集问题(无序)。接下来我们使用数据集进行案例演示:

in the algorithm foundation.there are tow kinds of algorithm are often used.one named full permutation problem,and another one named subSet problem.let us show it through a demostrate.

<1>数据:

elemnt:[1,2,3]

result:
1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 2 1 
3 1 2 

<2>数据:

elemnt:[1,2,3]

result:
1 2 3 
1 2 
1 3 
1 
2 3 
2 
3 

如何实现需求<1> <2>的需求?

<1>.1-----邻位交换法

<1>.1-----Ortho transposition algorithm
//程序大意:
//通过递归交换内部元素进行全排列的生成

package Algorithm;


//这是配列问题的第一种解法
//也成为邻位对换算法
public class ArrangementProblemsA1 {
    public static void main(String[] args) {
        int[] list = {1,2,3};
        Perm(list,0,list.length);
    }

    public static void Perm(int[] list, int k, int m) {
        //只剩下最后一个元素不需要置换
        if (k == m) {
            for(int index = 0;index<list.length;index++){
                System.out.print(list[index]+" ");
            }
            System.out.println();
        }
        else{
            for(int index = k;index<m;index++){
                //将list[k]作为首部
                Swap(list,index,k);
                //将list[k+1:m]作为递归条件
                Perm(list,k+1,m);
                //复位
                Swap(list,index,k);
            }
        }
    }

    public static void Swap(int[] list,int i,int j){
        int temp = list[i];
        list[i] = list[j];
        list[j] = temp;
    }
}

<1>.2-----label algorithm

<1>.2-----标签法

//程序大意:
//通过判断是否fang'wen

package Algorithm;


//这是配列问题的第二种解法
public class ArrangementProblemsA2 {

    static int n = 10;
    //设置结果数组
    static int[] result = new int[n];
    //设置标志位
    //默认为false
    static boolean[] hashTable = new boolean[n];
    static int[] list = {1, 2, 3, 4, 5};


    public static void main(String[] args) {

        //从index = 0开始全排列
        Perm(0);
    }

    public static void Perm(int index) {
        //如果排到最后一位的时候则可以直接打印数据
        if (index == 2) {

            for (int i = 0; i < list.length; i++) {
                System.out.print(result[i] + " ");
            }
            System.out.println();
        } else {
            for (int x = 0; x < list.length; x++) {
                //System.out.println(">>>>>>>>>>>>>>>>>"+index+"   "+list.length);
                //若当前标志位还在则进行置换
                if (hashTable[x] == false) {
                    result[index] = list[x];
                    hashTable[x] = true;
                    Perm(index+1);
                    //还原标志位
                    hashTable[x] = false;
                }
            }
        }
    }
}

<2>-----selective splicing algorithm

<2>-----选择拼接法

package Algorithm.Recursion;

import java.util.ArrayList;

/**
 * @author 韦海涛
 * @version 1.0
 * @date 4/17/2021 2:08 AM
 */
public class ArrangementProblemsC {
    //用于存放求取子集中的元素
    public static ArrayList<Integer> list = new ArrayList<Integer>();
    public static void getSubSet(int[] A, int step,int len) {
        while (step != A.length) {
            //递归执行语句,向数组链表中添加一个元素
            list.add(A[step]);
            step++;
            getSubSet(A, step,len);
            //回溯执行语句,删除数组链表最后一个元素
            list.remove(list.size() - 1);
        }
        for(int i = 0;i < list.size();i++) {
            System.out.print(list.get(i)+" ");
        }
        System.out.println();

    }

    public static void main(String[] args) {
        int [] arr = {1,2,3};
        getSubSet(arr,0,3);
    }
}

 

Sure, I can help you with that! Here's a C++ program that uses recursion and backtracking to find the correct lock combination: ```cpp #include <iostream> #include <vector> #include <cstdlib> #include <ctime> using namespace std; const int NUM_DIGITS = 5; void generateRandomCombination(vector<int>& combination) { for (int i = 0; i < NUM_DIGITS; i++) { combination.push_back(rand() % 10); } } void printCombination(const vector<int>& combination) { for (int i = 0; i < NUM_DIGITS; i++) { cout << combination[i]; } cout << endl; } bool tryCombination(vector<int>& currentCombination, const vector<int>& targetCombination, int digitIndex) { if (digitIndex == NUM_DIGITS) { return currentCombination == targetCombination; } for (int i = 0; i < 10; i++) { currentCombination[digitIndex] = i; if (tryCombination(currentCombination, targetCombination, digitIndex + 1)) { return true; } } return false; } int main() { srand(time(0)); vector<int> targetCombination; generateRandomCombination(targetCombination); cout << "Target combination: "; printCombination(targetCombination); vector<int> currentCombination(NUM_DIGITS, 0); if (tryCombination(currentCombination, targetCombination, 0)) { cout << "Found the correct combination: "; printCombination(currentCombination); } else { cout << "Unable to find the correct combination." << endl; } return 0; } ``` The `generateRandomCombination` function generates a random lock combination and stores it in a vector. The `printCombination` function simply prints out a given combination. The `tryCombination` function takes in the current combination being tested, the target combination to be found, and the current digit index that is being tested. It recursively tries all possible combinations of digits for the current index, and returns true if the correct combination is found. In the `main` function, we first generate a random target combination, print it out, and initialize the current combination to all zeros. We then call `tryCombination` with the current combination, target combination, and starting index of 0. If the correct combination is found, we print it out. Otherwise, we print a message saying that we were unable to find the correct combination. Here are three example recursive level calls for the recursion tree call with input values: ``` tryCombination({0, 0, 0, 0, 0}, {3, 2, 5, 8, 1}, 0) ├──tryCombination({0, 0, 0, 0, 3}, {3, 2, 5, 8, 1}, 1) │ ├──tryCombination({0, 0, 0, 2, 3}, {3, 2, 5, 8, 1}, 2) │ │ ├──tryCombination({0, 0, 5, 2, 3}, {3, 2, 5, 8, 1}, 3) │ │ │ ├──tryCombination({3, 0, 5, 2, 3}, {3, 2, 5, 8, 1}, 4) // returns true │ │ │ └──tryCombination({4, 0, 5, 2, 3}, {3, 2, 5, 8, 1}, 4) │ │ └──tryCombination({0, 1, 5, 2, 3}, {3, 2, 5, 8, 1}, 3) │ │ ├──tryCombination({3, 1, 5, 2, 3}, {3, 2, 5, 8, 1}, 4) // returns true │ │ └──tryCombination({4, 1, 5, 2, 3}, {3, 2, 5, 8, 1}, 4) │ └──tryCombination({0, 2, 5, 2, 3}, {3, 2, 5, 8, 1}, 3) │ ├──tryCombination({3, 2, 5, 2, 3}, {3, 2, 5, 8, 1}, 4) // returns true │ └──tryCombination({4, 2, 5, 2, 3}, {3, 2, 5, 8, 1}, 4) ├──tryCombination({0, 0, 0, 0, 4}, {3, 2, 5, 8, 1}, 1) │ ├──tryCombination({0, 0, 0, 2, 4}, {3, 2, 5, 8, 1}, 2) │ │ ├──tryCombination({0, 0, 5, 2, 4}, {3, 2, 5, 8, 1}, 3) │ │ │ ├──tryCombination({3, 0, 5, 2, 4}, {3, 2, 5, 8, 1}, 4) // returns true │ │ │ └──tryCombination({4, 0, 5, 2, 4}, {3, 2, 5, 8, 1}, 4) │ │ └──tryCombination({0, 1, 5, 2, 4}, {3, 2, 5, 8, 1}, 3) │ │ ├──tryCombination({3, 1, 5, 2, 4}, {3, 2, 5, 8, 1}, 4) // returns true │ │ └──tryCombination({4, 1, 5, 2, 4}, {3, 2, 5, 8, 1}, 4) │ └──tryCombination({0, 2, 5, 2, 4}, {3, 2, 5, 8, 1}, 3) │ ├──tryCombination({3, 2, 5, 2, 4}, {3, 2, 5, 8, 1}, 4) // returns true │ └──tryCombination({4, 2, 5, 2, 4}, {3, 2, 5, 8, 1}, 4) └──tryCombination({0, 0, 0, 0, 5}, {3, 2, 5, 8, 1}, 1) ├──tryCombination({0, 0, 0, 2, 5}, {3, 2, 5, 8, 1}, 2) │ ├──tryCombination({0, 0, 5, 2, 5}, {3, 2, 5, 8, 1}, 3) │ │ ├──tryCombination({3, 0, 5, 2, 5}, {3, 2, 5, 8, 1}, 4) // returns true │ │ └──tryCombination({4, 0, 5, 2, 5}, {3, 2, 5, 8, 1}, 4) │ └──tryCombination({0, 1, 5, 2, 5}, {3, 2, 5, 8, 1}, 3) │ ├──tryCombination({3, 1, 5, 2, 5}, {3, 2, 5, 8, 1}, 4) // returns true │ └──tryCombination({4, 1, 5, 2, 5}, {3, 2, 5, 8, 1}, 4) └──tryCombination({0, 2, 5, 2, 5}, {3, 2, 5, 8, 1}, 3) ├──tryCombination({3, 2, 5, 2, 5}, {3, 2, 5, 8, 1}, 4) // returns true └──tryCombination({4, 2, 5, 2, 5}, {3, 2, 5, 8, 1}, 4) ``` In this example, we're trying to find the correct combination for the lock with the target combination {3, 2, 5, 8, 1}. The recursion tree shows the different combinations being tried at each recursive call. At each level, the function tries all possible digits for the current index, and recursively calls itself with the next digit index. If a combination is found that matches the target combination, the function returns true and stops recursing.
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