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★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ Given a collection of distinct integers, return all possible permutations. Example: Input: [1,2,3] Output: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ] 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] ] 12ms
1 class Solution { 2 func permute (_ arr:[Int]) -> [[Int]] { 3 4 var array = arr 5 var result = [[Int]]() 6 helper(&array,0,&result) 7 return result 8 } 9 10 func helper (_ arr: inout [Int], _ begin:Int, _ results: inout [[Int]]){ 11 12 if begin >= arr.count{ 13 results.append(arr) 14 return 15 } else { 16 for i in begin..<arr.count{ 17 // choose 18 arr.swapAt(begin, i) 19 // explore 20 helper(&arr, begin + 1, &results) 21 // unchoose 22 arr.swapAt(begin, i) 23 } 24 } 25 } 26 } 16ms 1 class Solution { 2 func permute(_ nums: [Int]) -> [[Int]] { 3 4 var result = [[Int]]() 5 var temp = [Int]() 6 7 var isVisited : [Bool] = Array(repeating: false, count: nums.count) 8 backTracking(nums, &result, &temp, &isVisited) 9 return result 10 11 } 12 13 func backTracking(_ nums: [Int], _ result: inout [[Int]], _ temp: inout [Int], _ isVisited: inout [Bool]) { 14 if temp.count == nums.count { 15 result.append(temp) 16 return 17 } 18 for i in 0 ..< nums.count where !isVisited[i] { 19 temp.append(nums[i]) 20 isVisited[i] = true 21 backTracking(nums, &result, &temp, &isVisited) 22 isVisited[i] = false 23 temp.removeLast() 24 } 25 } 26 } 24ms 1 class Solution { 2 func permute(_ nums: [Int]) -> [[Int]] { 3 var result = [[Int]]() 4 var nums = nums 5 6 func permute(nums: inout [Int], start: Int, end: Int) { 7 if start == end { 8 result.append(nums) 9 } else { 10 for i in start...end { 11 nums.swapAt(i, start) 12 permute(nums: &nums, start: start + 1, end: end) 13 nums.swapAt(i, start) 14 } 15 } 16 } 17 permute(nums: &nums, start: 0, end: nums.count - 1) 18 return result 19 } 20 } 36ms 1 class Solution { 2 func permute(_ nums: [Int]) -> [[Int]] { 3 var result : [[Int]] = [] 4 permuteNext(from: nums, to: [], result: &result) 5 return result 6 } 7 func permuteNext(from : [Int] , to : [Int] , result : inout [[Int]]) { 8 if from.count == 0 { 9 result.append(to) 10 return 11 } 12 13 for idx in 0..<from.count { 14 var newfrom = from 15 newfrom.remove(at: idx) 16 permuteNext(from: newfrom, to: to + [from[idx]], result: &result) 17 } 18 } 19 } 36ms 1 class Solution { 2 var collectList: [[Int]] = [] 3 func permute(_ nums: [Int]) -> [[Int]] { 4 5 if(nums.count == 0) { 6 return collectList 7 } 8 9 permuteDFS(nums, start: 0, tempList: [Int]()) 10 11 return collectList 12 } 13 14 func permuteDFS(_ nums: [Int], start: Int, tempList: [Int]) { 15 16 var temp = tempList 17 18 if(temp.count == nums.count) { 19 20 collectList.append(temp) 21 22 }else { 23 24 for i in 0 ..< nums.count { 25 26 if( temp.contains(nums[i])) { 27 continue 28 } 29 30 temp.append(nums[i]) 31 32 permuteDFS(nums, start: (i + 1), tempList: temp) 33 34 temp.removeLast() 35 } 36 37 } 38 39 } 40 41 } 60ms 1 class Solution { 2 func permute(_ nums: [Int]) -> [[Int]] { 3 if nums.isEmpty { 4 return[[]] 5 } else if nums.count == 1 { 6 return [nums] 7 } else { 8 var result = [[Int]]() 9 for i in 0..<nums.count { 10 var nums = nums 11 let num = nums.remove(at: i) 12 for j in permute(nums) { 13 result.append([num] + j) 14 } 15 } 16 return result 17 } 18 } 19 } 80ms 1 class Solution { 2 func permute(_ nums: [Int]) -> [[Int]] { 3 var result: [[Int]] = [] 4 permute(nums, [], &result) 5 return result 6 } 7 8 func permute(_ nums: [Int], _ nums2: [Int], _ result: inout [[Int]]) { 9 guard nums2.count != nums.count else { 10 result.append(nums2) 11 return 12 } 13 14 nums.filter { !nums2.contains($0) }.forEach { (num) in 15 var nums4 = nums2 16 nums4.append(num) 17 permute(nums, nums4, &result) 18 } 19 } 20 }
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