47. Permutations II

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

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

难度：medium

题目：给定一组可能包含重复数的集合，返回所有可能的排列。

思路：先对数组进行排序，为去重做准备。然后借助递归遍历所有数。重点在于，当每次选定一个数做完交换之后，恢复时要对两个交换数及之间的所有数做排序。继续为接下来的去重做准备。

class Solution {public List<List<Integer>> permuteUnique(int[] nums) {Arrays.sort(nums);List<List<Integer>> result = new ArrayList<>();permuteUnique(nums, 0, new Stack<Integer>(), result);return result;}private void permuteUnique(int[] nums, int idx, Stack<Integer> stack, List<List<Integer>> result) {if (idx == nums.length) {result.add(new ArrayList<>(stack));return;}for (int i = idx; i < nums.length; i++) {if (idx == i || nums[i] != nums[i - 1]) {stack.push(nums[i]);swap(nums, idx, i);permuteUnique(nums, idx + 1, stack, result);swap(nums, i, idx);Arrays.sort(nums, idx, i + 1);stack.pop();}}}private void swap(int[] nums, int i, int j) {int t = nums[i];nums[i] = nums[j];nums[j] = t;}
}

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