Permutations II, Leetcode 解题笔记

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

只是在前一题的基础上去重,很简单的修改:

public class Solution {
    
    public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
        ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> sol = new ArrayList<Integer>();
        Arrays.sort(num); //这里不要忘了先排序
        int[] visited = new int[num.length];
        dfs(ret, sol, num, visited, 0);
        return ret;
    }
    
    public void dfs(ArrayList<ArrayList<Integer>> ret, ArrayList<Integer> sol, int[] num, int[] visited, int deep){
        if(deep == num.length){
            ret.add(new ArrayList<Integer>(sol));
            return;
        }
        for(int i = 0; i < num.length; i++){
            if(visited[i] == 0){
                visited[i] = 1;
                sol.add(num[i]);
                dfs(ret, sol, num, visited, deep+1);
                sol.remove(sol.size()-1);
                visited[i] = 0;
                while(i < num.length-1 && num[i+1] == num[i]) i++; //如果当前数字与前一个重复,直接跳过
            }
        }
    }
}
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