LeetCode——Permutations II

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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].

原题链接：https://oj.leetcode.com/problems/permutations-ii/

题目：给定一组包含重复元素的集合，返回所有可能的独立排列。

思路：此题只要将之前的全排列那题稍作修改即可，首先将数组排序，这样，相同的元素就是相邻的了，在排列取元素时，如果发现前后元素相同，则跳过相同的值。

public List<List<Integer>> permuteUnique(int[] num) {if (num == null)return null;List<List<Integer>> result = new ArrayList<List<Integer>>();if (num.length == 0)return result;Arrays.sort(num);permute(num, new boolean[num.length], new ArrayList<Integer>(), result);return result;}public void permute(int[] num, boolean[] isused,ArrayList<Integer> current, List<List<Integer>> result) {if (current.size() == num.length) {result.add(new ArrayList<Integer>(current));return;}for (int i = 0; i < num.length; i++) {if (!isused[i]) {isused[i] = true;current.add(num[i]);permute(num, isused, current, result);isused[i] = false;current.remove(current.size() - 1);while (i + 1 < num.length && num[i + 1] == num[i])i++;}}}

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