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  • Permutations of Array

    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:https://www.cnblogs.com/midan/p/4703653.html
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