Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order.
方法：
回溯法，去重
boolean[] vis; public List<List<Integer>> permuteUnique(int[] nums) { List<List<Integer>> ans = new ArrayList<List<Integer>>(); List<Integer> perm = new ArrayList<Integer>(); vis = new boolean[nums.length]; Arrays.sort(nums); backtrack(nums, ans, 0, perm); return ans; } public void backtrack(int[] nums, List<List<Integer>> ans, int idx, List<Integer> perm) { if (idx == nums.length) { ans.add(new ArrayList<Integer>(perm)); return; } for (int i = 0; i < nums.length; ++i) { if (vis[i] || (i > 0 && nums[i] == nums[i - 1] && !vis[i - 1])) { continue; } perm.add(nums[i]); vis[i] = true; backtrack(nums, ans, idx + 1, perm); vis[i] = false; perm.remove(idx); } }
参考链接：
https://leetcode.com/problems/permutations-ii/
https://leetcode-cn.com/problems/permutations-ii/
原文：https://www.cnblogs.com/diameter/p/14178151.html