LeetCode 047 Permutations II

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题目描述

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].

分析

使用LeetCode 046 Permutations的方法，只是把rt变成Set，最后转成List，但是会超时。

参考自：http://leetcode.tgic.me/permutations-ii/index.html

代码

    static void swap(int x[], int a, int b) {        int t = x[a];        x[a] = x[b];        x[b] = t;    }    public static boolean nextPermutation(int[] num) {        if (num.length < 2)            return false;        int p = -1;        for (int i = num.length - 1; i > 0; i--) {            if (num[i] > num[i - 1]) {                p = i - 1;                break;            }        }        if (p == -1) {            Arrays.sort(num);            return false;        }        int c = -1;        for (int i = num.length - 1; i >= 0; i--) {            if (num[i] > num[p]) {                c = i;                break;            }        }        swap(num, p, c);        Arrays.sort(num, p + 1, num.length);        return true;    }    List<Integer> asList(int[] num) {        ArrayList<Integer> l = new ArrayList<Integer>(num.length);        for (int i = 0; i < num.length; i++)            l.add(num[i]);        return l;    }    public List<List<Integer>> permuteUnique(int[] num) {        Arrays.sort(num);        ArrayList<List<Integer>> found = new ArrayList<List<Integer>>();        found.add(asList(num));        while (nextPermutation(num)) {            found.add(asList(num));        }        return found;    }

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