给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。
示例 1：
输入：nums = [1,1,2]
输出：
[[1,1,2],
[1,2,1],
[2,1,1]]示例 2：
输入：nums = [1,2,3]
输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]提示：
- 1 <= nums.length <= 8
- -10 <= nums[i] <= 10
解答：
昨天的代码改改就好了，增加了排序，遇到重复的数就跳过
func permuteUnique(nums []int) [][]int {
l := len(nums)
res := make([][]int, 0)
if l == 0 {
return res
}
sort.Ints(nums)
tmp := make([]int, 0)
used := make([]bool, l)
res = dfs(nums, l, 0, tmp, used, res)
return res
}
func dfs(nums []int, l, depth int, tmp []int, used []bool, res [][]int) [][]int {
if depth == l {
a := make([]int, l)
copy(a, tmp)
return append(res, a)
}
for i := 0; i < l; i++ {
if used[i] || i > 0 && !used[i-1] && nums[i] == nums[i-1] {
continue
}
tmp = append(tmp, nums[i])
used[i] = true
res = dfs(nums, l, depth+1, tmp, used, res)
tmp = tmp[:len(tmp)-1]
used[i] = false
}
return res
}官方解答：
1.搜索回溯
func permuteUnique(nums []int) (ans [][]int) {
sort.Ints(nums)
n := len(nums)
var perm []int
vis := make([]bool, n)
var backtrack func(int)
backtrack = func(idx int) {
if idx == n {
ans = append(ans, append([]int(nil), perm...))
return
}
for i, v := range nums {
if vis[i] || i > 0 && !vis[i-1] && v == nums[i-1] {
continue
}
perm = append(perm, v)
vis[i] = true
backtrack(idx + 1)
vis[i] = false
perm = perm[:len(perm)-1]
}
}
backtrack(0)
return
}