LeetCode_ 47_ Full arrangement II

Topic link

• https://leetcode.cn/problems/permutations-ii/

Title Description

Given a sequence that can contain repeating numbers nums , In any order Returns all non repeating permutations .

Example 1：

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

Example 2：

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

Tips ：

• 1 <= nums.length <= 8
• -10 <= nums[i] <= 10

Their thinking

Law 1 ： backtracking + No pruning

Is to traverse all possibilities , Then go heavy.

AC Code

Law 1

class Solution {
    
    boolean[] flag;
    List<Integer> path = new ArrayList<>();
    List<List<Integer>> ans = new ArrayList<>();

    public List<List<Integer>> permuteUnique(int[] nums) {
    
        flag = new boolean[nums.length];
        backTracing(nums);
        return ans;
    }

    private void backTracing(int[] nums) {
    
        if (path.size() == nums.length && !ans.contains(path)) {
    
            ans.add(new ArrayList<>(path));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
    
            if (flag[i]) {
    
                continue;
            }
            path.add(nums[i]);
            flag[i] = true;
            backTracing(nums);
            path.remove(path.size() - 1);
            flag[i] = false;
        }
    }
}