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], [2,1,1] ]
和一般的Permutation不一样的是,这种permutation需要排序,使相同的元素能够相邻,选取下一个元素的时候,要查看这个元素的前一个元素是否和它相同,如果相同而且没有使用过,就不用选取这个元素,因为如果选取了这个元素,所得的结果被包含于选取了前一个相同元素的结果中。
代码如下:
public class Solution { int length; public List<List<Integer>> permuteUnique(int[] nums) { length = nums.length; List<List<Integer>> result = new ArrayList<List<Integer>>(); if (length == 0) { return result; } Arrays.sort(nums); boolean[] flags = new boolean[length]; Arrays.fill(flags, false); List<Integer> candidate = new ArrayList<Integer>(); helper(nums, length, flags, result, candidate); return result; } private void helper(int[] nums, int n, boolean[] flags, List<List<Integer>> result, List<Integer> candidate) { if (n == 0) { result.add(new ArrayList<Integer>(candidate)); } int ll = candidate.size(); for (int i = 0; i < length; i++) { if (!flags[i]) { //if the number appears more than once and the same number before it has not been used if (i != 0 && nums[i] == nums[i - 1] && !flags[i - 1]) { continue; } flags[i] = true; candidate.add(nums[i]); helper(nums, n - 1, flags, result, candidate); candidate.remove(ll); flags[i] = false; } } } }