LeetCode 第 46 题：全排列（回溯）

地址：https://leetcode-cn.com/problems/permutations/

我写的题解地址：https://leetcode-cn.com/problems/permutations/solution/hui-su-suan-fa-python-dai-ma-java-dai-ma-by-liweiw/

Java 代码：

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;

public class Solution {

    // 时间复杂度: O(n^n)
    // 空间复杂度: O(n)

    public List<List<Integer>> permute(int[] nums) {
        int len = nums.length;

        List<List<Integer>> res = new ArrayList<>(len);
        if (len == 0) {
            return res;
        }

        boolean[] used = new boolean[len];
        Deque<Integer> stack = new ArrayDeque<>(len);

        backtrack(nums, 0, len, used, stack, res);
        return res;
    }

    private void backtrack(int[] nums, int index, int len, boolean[] used, Deque<Integer> stack, List<List<Integer>> res) {
        if (index == len) {
            res.add(new ArrayList<>(stack));
            return;
        }

        for (int i = 0; i < len; i++) {
            if (!used[i]) {
                used[i] = true;
                stack.addLast(nums[i]);

                backtrack(nums, index + 1, len, used, stack, res);

                stack.removeLast();
                used[i] = false;
            }
        }
    }
}

复杂度分析：

• 时间复杂度： $O(N^N)$
• 空间复杂度： $O(N)$

说明：

1、这里的 used 数组记录了当前所在结点的状态；

2、used 数组可以使用哈希表或者“位掩码”替代。

Java 代码：

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class Solution {

    public List<List<Integer>> permute(int[] nums) {
        int len = nums.length;

        List<List<Integer>> res = new ArrayList<>(len);
        if (len == 0) {
            return res;
        }

        // 使用哈希表检测一个数字是否使用过
        Set<Integer> used = new HashSet<>(len);
        Deque<Integer> stack = new ArrayDeque<>(len);

        backtrack(nums, 0, len, used, stack, res);
        return res;
    }

    private void backtrack(int[] nums, int index, int len, Set<Integer> used, Deque<Integer> stack, List<List<Integer>> res) {
        if (index == len) {
            res.add(new ArrayList<>(stack));
            return;
        }

        for (int i = 0; i < len; i++) {
            if (!used.contains(i)) {
                used.add(i);
                stack.addLast(nums[i]);

                backtrack(nums, index + 1, len, used, stack, res);

                stack.removeLast();
                used.remove(i);
            }
        }
    }
}

Java 代码：

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;


public class Solution {

    public List<List<Integer>> permute(int[] nums) {
        int len = nums.length;
        List<List<Integer>> res = new ArrayList<>(len);
        if (len == 0) {
            return res;
        }

        int hash = 0;
        Deque<Integer> stack = new ArrayDeque<>(len);
        backtracking(nums, 0, hash, stack, len, res);
        return res;
    }

    private void backtracking(int[] nums, int index, int hash, Deque<Integer> stack, int len, List<List<Integer>> res) {
        if (index == len) {
            res.add(new ArrayList<>(stack));
            return;
        }

        for (int i = 0; i < len; i++) {
            if (((hash >> i) & 1) == 0) {
                stack.addLast(nums[i]);
                hash ^= (1 << i);

                backtracking(nums, index + 1, hash, stack, len, res);
                hash ^= (1 << i);
                stack.removeLast();
            }
        }
    }
}