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排列与组合

组合:

public class Solution {
    static int  count = 0;
    static void findSo(String soFar,String rest) {
        if(soFar.length() == 3){
            count++;
            System.out.println(soFar);
        }else{
            for(int i=0; i<rest.length(); i++){
                String now;
                if(soFar.length()<3){
                  now = soFar+rest.charAt(i);
                }else{
                  now = soFar;
                }
                String remain = rest.substring(i+1);
                findSo(now,remain);
            }
            
        }
    }

    public static void main(String[] args) {
        String str  ="1234567";
        findSo("",str);
        System.out.println("count:"+count);
    }
}

结果:

C++版本

#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;
int count;
void findSo(string soFar, string rest) {
        if (soFar.length() == 3) {
            count++;
            cout<<soFar<<endl;
        } else {
            for (int i = 0; i < rest.length(); i++) {
                string now;
                if(soFar.length()<3){
                    now = soFar + rest.at(i);
                }else{
                    now  = soFar;
                }
                string remain = rest.substr(i + 1);
                findSo(now, remain);
            }

        }
}
int main()
{
    string str = "1234567";
    findSo("", str);
    cout<<"count:"<<count<<endl;
}

排列:

public class Solution {
    static int count = 0;

    static void findSo(String soFar, String rest) {
        if (soFar.length() == 3) {
            count++;
            System.out.println(soFar);
        } else {
            for (int i = 0; i < rest.length(); i++) {
                String now = soFar + rest.charAt(i);
                String remain = rest.substring(0,i) + rest.substring(i + 1);
                findSo(now, remain);
            }

        }
    }

    public static void main(String[] args) {
        String str = "1234567";
        findSo("", str);
        System.out.println("count:" + count);
    }
}

for循环里面的代码也可以和组合一样

　　　　　　　for (int i = 0; i < rest.length(); i++) {
                String now;
                if (soFar.length() < 3) {
                    now = soFar + rest.charAt(i);
                } else {
                    now = soFar;
                }
                String remain = rest.substring(0, i) + rest.substring(i + 1);
                findSo(now, remain);
            }

结果:

View Code

C++版本:

#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;
int count;
void findSo(string soFar, string rest) {
        if (soFar.length() == 3) {
            count++;
            cout<<soFar<<endl;
        } else {
            for (int i = 0; i < rest.length(); i++) {
                string now = soFar + rest.at(i);
                string remain = rest.substr(0,i) + rest.substr(i + 1);
                findSo(now, remain);
            }

        }
}
int main()
{
    string str = "1234567";
    findSo("", str);
    cout<<"count:"<<count<<endl;
}

从 m个元素中去n个元素的排列和组合差别仅仅在于

排列 string remain = rest.substr(0,i)+rest.substr(i + 1);

组合 string remain = rest.substr(i + 1);

所以组合代码也可以是:

#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;
int count;
void findSo(string soFar, string rest)
{
    if (soFar.length() == 3)
    {
        count++;
        cout<<soFar<<endl;
    }
    else
    {
        for (int i = 0; i < rest.length(); i++)
        {
            string  now = soFar + rest.at(i);
            string remain = rest.substr(i + 1);
            findSo(now, remain);
        }

    }
}
int main()
{
    string str = "1234567";
    findSo("", str);
    cout<<"count:"<<count<<endl;
}

ps:

抽象的想一下以1 2 3 4 5 6 7为例
for循环刚进来递归肯定会得到7个大分支
不好意思,他现在先要处理第一个分支
即("1","2,3,4,5,6,7) 在这个分支下不断的分
递归每得到一个分支就要穷尽这个分支
穷尽之后在返回到父节点的相邻分支继续穷尽

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原文地址：https://www.cnblogs.com/hixin/p/4788097.html