Permutations && Permutations ii

最新推荐文章于 2023-10-21 15:01:03 发布

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最新推荐文章于 2023-10-21 15:01:03 发布

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分类专栏： leetcode 文章标签：链表函数

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本文详细解析了Permutations算法（不含重复元素排列）与PermutationsII算法（含重复元素排列）的实现原理及Java代码实现。Permutations算法通过深度优先搜索实现全排列，而PermutationsII算法在此基础上增加了跳过重复元素的功能，确保结果中没有重复的排列。

Permutations 中不带有重复元素

Permutations II 中带有重复元素

Permutations

java

public class Solution {
    /*
     * @param nums: A list of integers.
     * @return: A list of permutations.
     */
    public List<List<Integer>> permute(int[] nums) {
        // write your code here
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            result.add(new ArrayList<Integer>());
            return result;
        }
        List<Integer> path = new ArrayList<>();
        Set<Integer> set = new HashSet<>();
        dfs(nums, path, result, set);
        return result;
    }
    private void dfs(int[] nums, 
                     List<Integer> path,
                     List<List<Integer>> result,
                     Set<Integer> set) {
        if (path.size() == nums.length) {
            result.add(new ArrayList<Integer>(path));
            return;
        }          
        for (int i = 0; i < nums.length; i++) {
            if (!set.contains(nums[i])) {
                set.add(nums[i]);
                path.add(nums[i]);
                dfs(nums, path, result, set);
                set.remove(nums[i]);
                path.remove(path.size() - 1);
            }
        }
    }
}

Permutations II

java

public class Solution {
    /*
     * @param :  A list of integers
     * @return: A list of unique permutations
     */
    public List<List<Integer>> permuteUnique(int[] nums) {
        // write your code here
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            result.add(new ArrayList<Integer>());
            return result;
        } 
        List<Integer> path = new ArrayList<>();
        boolean[] visit = new boolean[nums.length];
        Arrays.sort(nums);
        dfs(nums, path, result, visit);
        return result;
    }
    private void dfs(int[] nums, 
                     List<Integer> path, 
                     List<List<Integer>> result,
                     boolean[] visit) {
        if (path.size() == nums.length) {
            result.add(new ArrayList<Integer>(path));
            return;
        }          
        for (int i = 0; i < nums.length; i++) {
            if ((visit[i] == true) || (i != 0 && nums[i] == nums[i - 1] && visit[i - 1] == false)) {
                continue;
            } else {
                path.add(nums[i]);
                visit[i] = true;
                dfs(nums, path, result, visit);
                visit[i] = false;
                path.remove(path.size() - 1);
            }
        }
    }
}