LeetCode 46. Permutations

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Given a collection of distinct numbers,return all possible permutations.

For example,

[1,2,3] have the following permutations:

[

[1,2,3],

[1,3,2],

[2,1,3],

[2,3,1],

[3,1,2],

[3,2,1]

]

思路一：通过遍历线性表每次选择一个元素将其加入链表，删除该元素并递归执行，当线性表中没有元素是结束递归。

public class Node{                   intval;                   Nodenext;                   publicNode(int val){                            this.val= val;                   }         }public voiddfs(List<List<Integer>> res, Node head, List<Integer> ans){             if(head.next==null) {res.add(newArrayList<Integer>(ans)); return;}             Node p = head, q=head.next;              while(q!=null){                       ans.add(q.val);                       p.next = q.next;                       dfs(res,head,ans);                       ans.remove(ans.size()-1);                       p.next = q;                       p = p.next; q= q.next;             }}

完整代码：

class Solution {         publicclass Node{                   intval;                   Nodenext;                   publicNode(int val){                            this.val= val;                   }         }   public List<List<Integer>> permute(int[] nums) {       List<List<Integer>> res = newArrayList<List<Integer>>();       if(nums.length == 0) return res;       Arrays.sort(nums);         Node head= new Node(0);       Node p=head;       for(int i=0;i<nums.length;i++){                Node tmp=newNode(nums[i]);                p.next = tmp;                p = p.next;       }       dfs(res,head,new ArrayList<Integer>());       return res;    }    public voiddfs(List<List<Integer>> res, Node head, List<Integer> ans){             if(head.next==null) {res.add(newArrayList<Integer>(ans)); return;}             Node p = head, q=head.next;              while(q!=null){                       ans.add(q.val);                       p.next = q.next;                       dfs(res,head,ans);                       ans.remove(ans.size()-1);                       p.next = q;                       p = p.next; q= q.next;             }    }}

思路二：通过每次交换当前元素i 与 i+k元素递归调用：

private void backtrack(int[]nums,List<List<Integer>> result,List<Integer> res,int index)    {       if(index==nums.length)       {            result.add(newArrayList<Integer>(res));           return;       }              for(int i=index;i<nums.length;i++)       {           swap(nums,index,i);           res.add(nums[index]);           backtrack(nums,result,res,index+1);           res.remove(res.size()-1);           swap(nums,index,i);       }}

完整代码：

class Solution {   public List<List<Integer>> permute(int[] nums) {              List<List<Integer>> result = new ArrayList<>();       List<Integer> res = new ArrayList<>();              if(nums!=null && nums.length>0)           backtrack(nums,result,res,0);              return result;           }      private void backtrack(int[] nums,List<List<Integer>>result,List<Integer> res,int index)    {       if(index==nums.length)       {           result.add(new ArrayList<Integer>(res));           return;       }              for(int i=index;i<nums.length;i++)       {           swap(nums,index,i);           res.add(nums[index]);            backtrack(nums,result,res,index+1);           res.remove(res.size()-1);           swap(nums,index,i);       }    }      private void swap(int[] nums,int i,int j)    {       int tmp = nums[i];       nums[i]=nums[j];       nums[j]=tmp;    }}

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