LeetCode——Permutations II

标签：leetcode

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].

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++;
			}
		}
	}

LeetCode——Permutations II

标签：leetcode

原文地址：http://blog.csdn.net/laozhaokun/article/details/38946865