46. 全排列 (opens new window)

English Version (opens new window)

题目描述

给定一个 没有重复 数字的序列，返回其所有可能的全排列。

示例:

输入: [1,2,3]
输出:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

解法

Python3

回溯法：

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        def dfs(nums, i, res, path, used):
            if i == len(nums):
                res.append(copy.deepcopy(path))
                return
            for j in range(len(nums)):
                if not used[j]:
                    path.append(nums[j])
                    used[j] = True
                    dfs(nums, i + 1, res, path, used)
                    used[j] = False
                    path.pop()

        res, path = [], []
        used = [False] * len(nums)
        dfs(nums, 0, res, path, used)
        return res

切分数组：

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        if len(nums) <= 1:
            return [nums]
        res = []
        for i, num in enumerate(nums):
            n = nums[:i] + nums[i + 1:]
            for item in self.permute(n):
                res.append([num] + item)
        return res

Java

回溯法：

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        boolean[] used = new boolean[nums.length];
        dfs(nums, 0, res, path, used);
        return res;
    }

    private void dfs(int[] nums, int i, List<List<Integer>> res, List<Integer> path, boolean[] used) {
        if (i == nums.length) {
            res.add(new ArrayList<>(path));
            return;
        }
        for (int j = 0; j < nums.length; ++j) {
            if (!used[j]) {
                path.add(nums[j]);
                used[j] = true;
                dfs(nums, i + 1, res, path, used);
                used[j] = false;
                path.remove(path.size() - 1);
            }
        }
    }
}

• 递归：

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        permute(res, nums, 0);
        return res;
    }

    private void permute(List<List<Integer>> res, int[] nums, int start) {
        if (start == nums.length) {
            List<Integer> t = new ArrayList<>();
            for (int e : nums) {
                t.add(e);
            }
            res.add(t);
            return;
        }
        for (int i = start; i < nums.length; ++i) {
            swap(nums, i, start);
            permute(res, nums, start + 1);
            swap(nums, i, start);
        }
    }

    private void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permute = function(nums) {
    let res = [];
    let solution = [];
    let record = new Array(nums.length).fill(false);
    dfs(nums, 0, record, solution, res);
    return res;
};

function dfs (nums, depth, record, solution, res) {
    if (depth == nums.length) {
        res.push(solution.slice());
        return;
    }
    for (let i = 0; i < nums.length; i++) {
        if (!record[i]) {
            solution.push(nums[i]);
            record[i] = true;
            dfs(nums, depth + 1, record, solution, res);
            solution.pop();
            record[i] = false;
        }
    }
}

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