47. Permutations II ⭐

Medium

Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order.

Example 1:

Input: nums = [1,1,2]
Output:
[[1,1,2],
 [1,2,1],
 [2,1,1]]

Example 2:

Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Constraints:

• 1 <= nums.length <= 8

• -10 <= nums[i] <= 10

解題

回圈內第一個 if 的 !visited[i - 1] 其實是代表該值已經被處理過，狀態回復了

func permuteUnique(nums []int) [][]int {
    sort.Ints(nums)

    ans := make([][]int, 0)
    perm := make([]int, len(nums))
    visited := make([]bool, len(nums)) // 一個狀態變量儲存整個狀態樹

    var backtracking func(int)
    backtracking = func(index int) {
        if index == len(nums) {
            ans = append(ans, append([]int{}, perm...))
            return
        }

        for i:=0; i<len(nums); i++ {
            if i > 0 && nums[i] == nums[i - 1] && !visited[i - 1] { 
                continue
            }
            if !visited[i] {
                visited[i] = true
                perm[index] = nums[i]
                backtracking(index+1)   
                visited[i] = false // 撤銷選擇
            }
        }
    }

    backtracking(0)

    return ans
}