题目:
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:
这道题是找所有的全排列,典型的排列组合模板,用dfs,因为nums里的数不重复,所以相对容易,下面程序中helper的作用是寻找所有以permute开头的全排列。
class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> res=new ArrayList<List<Integer>>();
List<Integer> permute=new ArrayList<Integer>();
Set<Integer> set=new HashSet<>();
if(nums==null)
return res;
Arrays.sort(nums);
helper(res,permute,nums,set);
return res;
}
public void helper(List<List<Integer>> res,List<Integer> permute,int[] nums,Set<Integer> set){
if(permute.size()==nums.length)
res.add(new ArrayList<Integer>(permute));
for(int i=0;i<nums.length;i++){
if(!set.contains(nums[i])){
set.add(nums[i]);
permute.add(nums[i]);
helper(res,permute,nums,set);
set.remove(nums[i]);
permute.remove(permute.size()-1);
}
}
}
}
下面这么写也可以:
for(int i=0;i<nums.length;i++){
if(set.contains(nums[i])){
continue;//跳过本次循环进行下次循环,而break是直接跳出循环
}
set.add(nums[i]);
permute.add(nums[i]);
helper(res,permute,nums,set);
set.remove(nums[i]);
permute.remove(permute.size()-1);
}
for循环一开始写成下面这样,是不对的:
for(int i=0;i<nums.length;i++){
if(!set.contains(nums[i])){
set.add(nums[i]);
permute.add(nums[i]);
}
helper(res,permute,nums,set);
set.remove(nums[i]);
permute.remove(permute.size()-1);
}
另外,set因为是无序的,remove(某个对象) 而list的remove方法是remove(对象的索引)