46. Permutations 全排列

Posted 2020-12-12 immiao0319

Given a collection of distinct integers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

List<Integer>的大小是size（）
permutations，remove的是最后一位。(permutations.size() - 1);
nums数组为空时，应该是[[]]的结果。需要results.add(new ArrayList<>(permutations));
回溯法helper的返回类型应该是void 

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> results = new ArrayList<>();
        List<Integer> permutation = new ArrayList<>();
        HashSet<Integer> set = new HashSet<>();
            
        //2种cc
        if (nums == null) {
            return null;
        }
        
        if (nums.length == 0) {
            results.add(new ArrayList<Integer>());
            return results;
        }
        
        helper(nums, permutation, set, results);
        
        return results;
    }
    
    public void helper(int[] nums, 
                       List<Integer> permutation,
                       HashSet<Integer> set, 
                       List<List<Integer>> results) {
        //cc
        if (permutation.size() == nums.length) {
            results.add(new ArrayList(permutation));
            return ;
        }
        
        for (int i = 0; i < nums.length; i++) {
            if (set.contains(nums[i])) {
                continue;
            }
            
            set.add(nums[i]);
            permutation.add(nums[i]);
            helper(nums, permutation, set, results);
            permutation.remove(permutation.size() - 1);
            set.remove(nums[i]);
        }
    }
}

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