46.全排列.md

46. 全排列

url

题目

给定一个 没有重复 数字的序列，返回其所有可能的全排列。

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]
]

方法

code

js

let permute = nums => {
    let ret = [];
    if (nums === null || nums.length === 0) return ret;
    let marked = Array(nums.length).fill(false);
    let dfs = (nums, list) => {
        if (list.length === nums.length) {
            ret.push(list.slice())
            return;
        }
        for (let i = 0; i < nums.length; i++) {
            if (marked[i]) continue;
            list.push(nums[i]);
            marked[i] = true;
            dfs(nums, list);
            marked[i] = false;
            list.pop();
        }
    }
    dfs(nums, []);
    return ret;
}

go

func permute(nums []int) [][]int {
	var res [][]int
	var dfs func(nums []int, list []int)
	if len(nums) == 0 {
		return res
	}
	marked := make([]bool, len(nums))
	dfs = func(nums []int, list []int) {
		if len(list) == len(nums) {
			res = append(res, append([]int{}, list...))
			return
		}
		for i := 0; i < len(nums); i++ {
			if marked[i] {
				continue
			}
			list = append(list, nums[i])
			marked[i] = true
			dfs(nums, list)
			marked[i] = false
			list = list[0:len(list) - 1]
		}
	}
	dfs(nums, []int{})
	return res
}

java

class Solution {
    List<List<Integer>> ret = new ArrayList<>();
    public List<List<Integer>> permute(int[] nums) {
        if (nums == null || nums.length == 0)
            return ret;
        dfs(nums, new boolean[nums.length], new ArrayList<Integer>());
        return ret;
    }
    public void dfs(int[] nums, boolean[] marked, ArrayList<Integer> list){
        if (list.size() == nums.length){
            ret.add(new ArrayList(list));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if(marked[i])
                continue;
            list.add(nums[i]);
            marked[i] = true;
            dfs(nums, marked, list);
            marked[i]  = false;
            list.remove(list.size() - 1);
        }
    }
}