LeetCode 45 Permutations

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Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

思路：使用字典序法，思路见http://blog.csdn.net/mlweixiao/article/details/38893507,这样不管有没有重复的数字都可以处理。

public class Solution {private void nextPermutation(int[] num) {int i;int cur = -1;int temp;// find the last increase sequencefor (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>> permute(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;}}

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