LeetCode 45 Permutations

Given a collection of numbers、return all possible permuttions.
For example，[1,2,3] have the follwing permuttions：[1,2,3]、  [1,3,2]、  [2,1,3]、  [2,3,1]、  [3,1,2]、and  [3,2,1].
考え方：辞書の序文を使って、考え方は分かります。http://blog.csdn.net/mlweixiao/article/details/38893507このように、重複した数字があるかどうかに関わらず、処理できます。

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> permute(int[] num) {

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

		ArrayList> result = new ArrayList>();

		for (;;) {
			List list = new ArrayList();
			//       list.addAll(Arrays.asList(num));
			//      Collection.addAll(list ,num);
			// Arrays.asList()   java.util.Arrays$ArrayList，
			//    ArrayList。Arrays$ArrayList ArrayList    AbstractList，
			//remove，add method AbstractList    throw
			// UnsupportedOperationException        。
			//
			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;
	}
}