Permutations II

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题目

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].

方法

和上一题的方法一样，利用nextPermutation的思想。http://blog.csdn.net/u010378705/article/details/31348875

private boolean isReverseOrder(int[] num) {for (int i = 0; i < num.length - 1; i++) {if (num[i] < num[i + 1]) {return false;}}return true;}    public List<List<Integer>> permuteUnique(int[] num) {    List<List<Integer>> list = new ArrayList<List<Integer>>();    if (num == null || num.length == 0) {    return list;    }    Arrays.sort(num);    while (!isReverseOrder(num)) {    List<Integer> subList = new ArrayList<Integer>();    for (int i = 0; i < num.length; i++) {    subList.add(num[i]);    }    list.add(subList);    nextPermutation(num);    }    List<Integer> subList = new ArrayList<Integer>();for (int i = 0; i < num.length; i++) {subList.add(num[i]);}list.add(subList);nextPermutation(num);    return list;    }    public void nextPermutation(int[] num) {        if (num != null && num.length != 0 && num.length != 1) {            int len = num.length;            int i = len;            for (; i > 1; i--) {                if (num[i - 1] > num[i - 2]) {                                int flag = 0;                                for (int k = i - 1; k < len; k++) {                if (num[k] <= num[i - 2]) {                flag = k - 1;                break;                }                }                if (flag == 0) {                flag = len - 1;                }                int temp = num[flag];                num[flag] = num[i - 2];                num[i - 2] = temp;                Arrays.sort(num, i - 1, len);                break;                }                          }        }    }

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