0047.全排列II:优化排版,补充Swift版本

This commit is contained in:
bqlin
2021-12-13 15:54:48 +08:00
parent 39a233f22f
commit a953eca234

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@ -97,14 +97,15 @@ public:
大家发现,去重最为关键的代码为:
```
```cpp
if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
continue;
}
```
**如果改成 `used[i - 1] == true` 也是正确的!**,去重代码如下:
```
```cpp
if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == true) {
continue;
}
@ -131,13 +132,13 @@ if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == true) {
## 总结
这道题其实还是用了我们之前讲过的去重思路,但有意思的是,去重的代码中,这么写:
```
```cpp
if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
continue;
}
```
和这么写:
```
```cpp
if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == true) {
continue;
}
@ -291,7 +292,36 @@ var permuteUnique = function (nums) {
```
### Swift
```swift
func permuteUnique(_ nums: [Int]) -> [[Int]] {
let nums = nums.sorted() // 先排序,以方便相邻元素去重
var result = [[Int]]()
var path = [Int]()
var used = [Bool](repeating: false, count: nums.count)
func backtracking() {
if path.count == nums.count {
result.append(path)
return
}
for i in 0 ..< nums.count {
// !used[i - 1]表示同一树层nums[i - 1]使用过,直接跳过,这一步很关键!
if i > 0, nums[i] == nums[i - 1], !used[i - 1] { continue }
if used[i] { continue }
used[i] = true
path.append(nums[i])
backtracking()
// 回溯
path.removeLast()
used[i] = false
}
}
backtracking()
return result
}
```
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>