* write down thinking

I can not work out this without Google's help.

steps below are copied from Wikipedia:

1. Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation.

1. Find the largest index l greater than k such that a[k] < a[l].

1. Swap the value of a[k] with that of a[l].

1. Reverse the sequence from a[k + 1] up to and including the final element a[n].

[Fisherlei's Image](http://fisherlei.blogspot.com/2012/12/leetcode-next-permutation.html) is a good illustration.

You can see an image that how this solution goes.

* write down thinking

With the help of[Next Permutation](../next-permutation), you can find all unique permutations from the first permutation.

* First permutation is the numbers in non-descending order

* Last permutation is the numbers in non-ascending order

so the code is like

publicclassSolution {

voidswap(intx[],inta,intb) {

intt =x[a];

x[a] =x[b];

returntrue;

}

ArrayList<Integer>asList(int[]num){

List<Integer>asList(int[]num){

ArrayList<Integer>l =newArrayList<Integer>(num.length);

for(inti =0;i <num.length;i++)

l.add(num[i]);

returnl;

}

publicArrayList<ArrayList<Integer>>permuteUnique(int[]num) {

publicList<List<Integer>>permuteUnique(int[]num) {

Arrays.sort(num);

ArrayList<ArrayList<Integer>>found =newArrayList<ArrayList<Integer>>();

ArrayList<List<Integer>>found =newArrayList<List<Integer>>();

found.add(asList(num));

while(nextPermutation(num)){

returnfound;

}

}

}

layout:solution

title:Next Permutation

date:2014-07-23 02:42:48 +0800

date:2014-08-31 00:42:19 +0800

{% assign leetcode_name = {{page.path | remove: '/index.md'}} %}

{% assign leetcode_readme = {{leetcode_name | append: '/README.md' | prepend: '_root/' }} %}

layout:solution

title:Permutations II

date:2014-07-23 02:42:48 +0800

date:2014-08-31 00:52:32 +0800

{% assign leetcode_name = {{page.path | remove: '/index.md'}} %}

{% assign leetcode_readme = {{leetcode_name | append: '/README.md' | prepend: '_root/' }} %}