Summary of permutation and combination problems: (Summary of retrospective method)

1. Given a sequence without repeated numbers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2 ],
  [3,2,1]
]

First think about the template:

      public void dfs(int k){
        if (k==nums.length){
            System.out.println(tmp);    //排列完毕 打印。
            return;
        }
        for(int i=0;i<nums.length;i++ ){
            tmp.set(k,nums[i]);          //从第0位开始依次赋值
            dfs(k+1);
        }
    }

Write the following code

import java.util.*;

class Solution {
    List<List<Integer>> res;
    List<Integer> tmp = new ArrayList<>();
    boolean[] used ;
    public List<List<Integer>> permute(int[] nums) {
        used = new boolean[nums.length];                    //初始化
        for(int i=0;i<used.length;i++){
            used[i] =false;
            tmp.add(0);
        }

        res = new ArrayList<>();
        dfs(0, nums);
        return res;
    }

    public void dfs(int k, int[] nums) {            //套模板
        if (k == nums.length) {
            res.add(new ArrayList<>(tmp));    //注意这里
        
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (!used[i]) {
              
                tmp.set(k,nums[i]);
                used[i] = true;
                dfs(k + 1, nums);
                used[i] = false;
        
            }
        }
    }

}

Read the solution to a problem back thinking : The main difference in the code is almost the same here :()

  // tmp.add(0); 注释掉

public void dfs(int k, int[] nums) {            //套模板
        if (k == nums.length) {
            res.add(new ArrayList<>(tmp));
            System.out.println("out" + tmp);
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (!used[i]) {
                tmp.add(nums[i]);            
                used[i] = true;
                dfs(k + 1, nums);
                used[i] = false;
                tmp.remove(tmp.size()-1);    //回溯 还原状态
            }
        }
    }

2. Full arrangement II

Given a sequence nums that can contain repeated numbers, return all non-repeated permutations in any order.

 

Example 1:

Input: nums = [1,1,2]
Output:
[[1,1,2],
 [1,2,1],
 [2,1,1]]

Thinking is different from the above,

Method 1: You can directly de-duplicate in res

Method 2: Sort first and pruning when backtracking. The solution can be referred to: backtracking pruning
 

            if (used[i])
                continue;
            if(i>0 &&nums[i]==nums[i-1]&&!used[i-1])  //如果相等，且前一个没有使用
                continue;

 

 

3、

Given two integers n and k , return all possible combinations of k numbers in 1... n .

Example:

输入: n = 4, k = 2
输出:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

public class Solution {
    List<List<Integer>> res;
    List<Integer> tmp = new ArrayList<>();

    public static void main(String[] args) {
        Solution s = new Solution();
        s.combine(4,2);
    }

    public List<List<Integer>> combine(int n, int kk) {
        res = new ArrayList<>();
        if (kk <= 0 || n < kk) {
            return res;
        }
        dfs(1,kk,n);
        return res;
    }

    public void dfs(int b,int kk,int n) {            //套模板
        if (tmp.size()==kk) {
            res.add(new ArrayList<>(tmp));
            System.out.print(tmp);
            return;
        }
        for (int i = b; i <=n; i++) {
                tmp.add(i);
                dfs(i+1,kk,n);            //这里的不同点是i+1  不是上面排列中的k [[1, 2][1, 3][1, 4][2, 3][2, 4][3, 4]]
                // 如果改成上面排列中的的k（也就是改成b+1） [[1, 2][1, 3][1, 4][2, 2][2, 3][2, 4][3, 2][3, 3][3, 4][4, 2][4, 3][4, 4]]
                // 改成b [[1, 1][1, 2][1, 3][1, 4][2, 1][2, 2][2, 3][2, 4][3, 1][3, 2][3, 3][3, 4][4, 1][4, 2][4, 3][4, 4]]
                tmp.remove(tmp.size()-1);
        }
    }

}