LeetCode 47 Permutations II

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本文介绍了解决含有重复元素集合的所有不重复全排列问题的两种方法。通过递归回溯及迭代扩展子集的方式，实现对重复元素的有效处理，确保结果中没有重复的排列。

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],
  [2,1,1]
]

方法一：递归

	public List<List<Integer>> permuteUnique(int[] nums) {
		List<List<Integer>> result = new ArrayList<>();
		Arrays.sort(nums);
		backtrack(result, new ArrayList<Integer>(), nums, new boolean[nums.length]);
		return result;
	}

	private static void backtrack(List<List<Integer>> result, ArrayList<Integer> tempList, int[]
			nums, boolean[] used) {
		if (tempList.size() == nums.length) {
			result.add(new ArrayList<>(tempList));
		} else {
			for (int i = 0; i < nums.length; i++) {
				/**only insert duplicate element
				 when the previous duplicate element has been inserted*/
				if (used[i] || i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) continue;
				used[i] = true;
				tempList.add(nums[i]);
				backtrack(result, tempList, nums, used);
				used[i] = false;
				tempList.remove(tempList.size() - 1);
			}
		}
	}

方法二：

	public List<List<Integer>> permuteUnique3(int[] nums) {
		List<List<Integer>> result = new ArrayList<>();
		result.add(new ArrayList<Integer>());
		for (int i = 0; i < nums.length; i++) {
			List<List<Integer>> newres = new ArrayList<>();
			for (List<Integer> k : result) {
				for (int j = 0; j <= k.size(); j++) {
					if (j>0 && k.get(j-1) == nums[i]) break;//注意这里的判断和break
					List<Integer> list = new ArrayList<>(k);
					list.add(j, nums[i]);
					newres.add(list);
				}
			}
			result = newres;
		}
		return result;
	}