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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].
Thoughts:
You take first element of an array (k=0) and exchange it with any element (i) of the array. Then you recursively apply permutation on array starting with second element. This way you get all permutations starting with i-th element. The tricky part is that after recursive call you must swap i-th element with first element back, otherwise you could get repeated values at the first spot. By swapping it back we restore order of elements (basically you do backtracking).
Code:
public List<List<Integer>> permute(int[] nums) { List<List<Integer>> lists = new ArrayList<List<Integer>>(); p(nums,0,lists); return lists; } public void p(int[] nums, int k,List<List<Integer>> lists){ if(k>=nums.length){ List<Integer> current = new ArrayList<Integer>(); for (int a : nums) { current.add(a); } lists.add(current); } else{ for(int i=k;i<nums.length;i++){ swap(nums, i, k); p(nums,k+1,lists); swap(nums,i,k); // BackTracking } } } public void swap (int[] arr, int m, int n){ int temp = arr[m]; arr[m]=arr[n]; arr[n]=temp; }
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原文地址：http://www.cnblogs.com/midan/p/4703653.html
