LeetCode 46 Permutations II

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].

思路，使用字典序法，与http://blog.csdn.net/mlweixiao/article/details/38897499 代码除了函数名换了以外，没有任何改动

public class Solution {
	private void nextPermutation(int[] num) {
		int i;
		int cur = -1;
		int temp;
		// find the last increase sequence
		for (i = num.length - 1; i >= 1; i--) {
			if (num[i] > num[i - 1]) {
				cur = i - 1;
				break;
			}
		}

		// if the increase sequence exists,
		// swap the cur and the last one(bigger than it)
		if (cur != -1) {
			for (i = num.length - 1; i > cur; i--) {
				if (num[i] > num[cur]) {
					temp = num[cur];
					num[cur] = num[i];
					num[i] = temp;
					break;
				}
			}
		}

		for (i = cur + 1; 2 * i <= cur + num.length - 1; i++) {
			temp = num[i];
			num[i] = num[num.length - i + cur];
			num[num.length - i + cur] = temp;
		}
	}

	@SuppressWarnings({ "rawtypes", "unchecked" })
	public List<List<Integer>> permuteUnique(int[] num) {

		Arrays.sort(num);
		int[] temparray = new int[num.length];
		System.arraycopy(num, 0, temparray, 0, num.length);

		ArrayList<List<Integer>> result = new ArrayList<List<Integer>>();

		for (;;) {
			List list = new ArrayList<Integer>();
			// 注意不能使用list.addAll(Arrays.asList(num));
			// 也不能使用Collection.addAll(list ,num);
			// Arrays.asList() 返回java.util.Arrays$ArrayList，
			// 而不是ArrayList。Arrays$ArrayList和ArrayList都是继承AbstractList，
			//remove，add等method在AbstractList中是默认throw
			// UnsupportedOperationException而且不作任何操作。
			//
			for (int i = 0; i < num.length; i++) {
				list.add(num[i]);
			}
			result.add(list);
			nextPermutation(num);
			if (Arrays.equals(num, temparray))
				break;
		}
		return result;
	}