[Swift]LeetCode15. 三數之和 | 3Sum
阿新 • • 發佈：2018-10-31
Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
給定一個包含 n 個整數的陣列 nums，判斷 nums 中是否存在三個元素 a，b，c ，使得 a + b + c = 0 ？找出所有滿足條件且不重複的三元組。
注意：答案中不可以包含重複的三元組。
例如, 給定陣列 nums = [-1, 0, 1, 2, -1, -4]， 滿足要求的三元組集合為： [ [-1, 0, 1], [-1, -1, 2] ]
116ms
1 class Solution { 2 func threeSum(_ nums: [Int]) -> [[Int]] { 3 guard nums.count > 2 else { return [] } 4 var solutions = [[Int]](); 5 let sorted = nums.sorted() { $0 < $1 } 6 let count = sorted.count 7 var i = 0 8 9while (i < count - 2) { 10 if (i == 0 || (i > 0 && sorted[i] != sorted[i - 1])) { 11 var left = i + 1 12 var right = count - 1 13 let num = sorted[i] 14 15 while (left < right) { 16 let currentSum = sorted[left] + sorted[right] + num 17 18 if (currentSum == 0) { 19 solutions.append([sorted[left], sorted[right], num]) 20 21 while (left < right && sorted[left] == sorted[left + 1]) { 22 left += 1 23 } 24 25 while (left < right && sorted[right] == sorted[right - 1]) { 26 right -= 1 27 } 28 left += 1 29 right -= 1 30 } else if (currentSum < 0) { 31 left += 1 32 } else { 33 right -= 1 34 } 35 } 36 } 37 38 i += 1 39 } 40 41 return solutions 42 } 43 }
120ms
1 class Solution { 2 func threeSum(_ nums: [Int]) -> [[Int]] { 3 guard nums.count > 2 else { 4 return [] 5 } 6 7 var results = [[Int]]() 8 let sortedNums = nums.sorted() 9 10 for i in 0..<sortedNums.count-1 { 11 if i > 0 && sortedNums[i] == sortedNums[i-1] { 12 continue 13 } 14 let target = 0 - sortedNums[i] 15 var low = i + 1 16 var high = nums.count - 1 17 18 while low < high { 19 let sum = sortedNums[low] + sortedNums[high] 20 if sum == target { 21 let result = [sortedNums[i], sortedNums[low], sortedNums[high]] 22 results.append(result) 23 24 while (low < high && sortedNums[low] == sortedNums[low+1]) { 25 low += 1 26 } 27 while (low < high && sortedNums[high] == sortedNums[high-1]) { 28 high -= 1 29 } 30 low += 1 31 high -= 1 32 } else if sum < target { 33 low += 1 34 } else { 35 high -= 1 36 } 37 } 38 } 39 40 return results 41 } 42 }
124ms
1 class Solution { 2 func threeSum(_ nums: [Int]) -> [[Int]] { 3 if nums.count < 3 { 4 return [] 5 } 6 7 var result = [[Int]]() 8 var snums = nums.sorted() 9 10 for i in 0...snums.count-2 { 11 if i > 0 && snums[i] == snums[i-1] { 12 continue 13 } 14 15 var l = i + 1 16 var r = snums.count - 1 17 18 while l < r { 19 let s = snums[i] + snums[l] + snums[r] 20 21 if s == 0 { 22 result.append([snums[i], snums[l], snums[r]]) 23 l += 1 24 r -= 1 25 26 while l < r && snums[l] == snums[l-1] { 27 l += 1 28 } 29 30 while l < r && snums[r] == snums[r+1] { 31 r -= 1 32 } 33 } else if s < 0 { 34 l += 1 35 } else { 36 r -= 1 37 } 38 } 39 } 40 41 return result 42 } 43 }
136ms
1 class Solution { 2 func threeSum(_ nums: [Int]) -> [[Int]] { 3 var nums = nums.sorted() 4 var result = [[Int]]() 5 6 for i in 0..<nums.count { 7 if i == 0 || i > 0 && nums[i] != nums[i-1] { 8 var left = i + 1, right = nums.count - 1 9 var sum = 0 - nums[i] 10 print(sum) 11 while left < right { 12 if nums[left] + nums[right] == sum { 13 result.append([nums[left], nums[right], nums[i]]) 14 while left < right && nums[left] == nums[left + 1]{ 15 left += 1 16 } 17 while left < right && nums[right] == nums[right - 1]{ 18 right -= 1 19 } 20 left += 1 21 right -= 1 22 } else if nums[left] + nums[right] < sum { 23 left += 1 24 } else { 25 right -= 1 26 } 27 } 28 } 29 } 30 return result 31 } 32 }
148ms
1 import Foundation 2 class Solution { 3 func threeSum(_ nums: [Int]) -> [[Int]] { 4 5 var n = nums 6 n.sort() 7 8 var aIndex1: Int = 0 9 var aIndexLow: Int = 0 10 var aIndexHi: Int = 0 11 12 var aResult = [[Int]]() 13 14 var aTargetSum: Int = 0 15 16 while aIndex1 < (n.count - 2) { 17 18 //The sandwich principle. 19 aIndexLow = aIndex1 + 1 20 aIndexHi = n.count - 1 21 22 while (aIndexLow < aIndexHi) && (n[aIndex1] + n[aIndexLow]) <= aTargetSum { 23 24 var aSum = n[aIndex1] + n[aIndexLow] + n[aIndexHi] 25 if aSum == aTargetSum { 26 var aArr = [n[aIndex1], n[aIndexLow], n[aIndexHi]] 27 aResult.append(aArr) 28 29 aIndexHi -= 1 30 //Prevent dupes on hi 31 while aIndexLow < aIndexHi && n[aIndexHi + 1] == n[aIndexHi] { 32 aIndexHi -= 1 33 } 34 35 //Prevent dupes on low 36 aIndexLow += 1 37 while aIndexLow < aIndexHi && n[aIndexLow - 1] == n[aIndexLow] { 38 aIndexLow += 1 39 } 40 } else if aSum > aTargetSum { 41 aIndexHi -= 1 42 } else { 43 aIndexLow += 1 44 } 45 } 46 47 //prevent dupes with first index. 48 aIndex1 += 1 49 while aIndex1 < n.count && n[aIndex1 - 1] == n[aIndex1] { 50 aIndex1 += 1 51 } 52 } 53 54 return aResult 55 } 56 }
164ms
1 class Solution { 2 func twoSum(_ array: [Int], ignoreIndex i: Int, result: inout [[Int]]) { 3 4 var l = i + 1 5 6 let a = array[i] 7 8 var r = array.count - 1 9 10 if i > 0 && a == array[i - 1] { 11 return 12 } 13 14 while l < r { 15 16 let b = array[l] 17 18 if a + b > 0 { 19 return 20 } 21 22 let c = array[r] 23 24 var sum = a + b + c 25 26 if sum == 0 { 27 let indexes = [a, b, c] 28 29 result.append(indexes) 30 31 l += 1 32 33 while l < r && array[l] == b { 34 l += 1 35 } 36 37 } else if sum < 0 { 38 l += 1 39 40 while l < r && array[l] == b { 41 l += 1 42 } 43 } else if sum > 0 { 44 r -= 1 45 46 while l < r && array[r] == c { 47 r -= 1 48 } 49 } 50 } 51 52 } 53 54 func threeSum(_ nums: [Int]) -> [[Int]] { 55 56 var result = [[Int]]() 57 58 if nums.count < 3 { 59 return result 60 } 61 62 let array = nums.sorted { 63 return $0 < $1 64 } 65 66 for i in 0..<array.count - 2 { 67 if array[i] > 0 { 68 break 69 } 70 71 twoSum(array, ignoreIndex: i, result: &result) 72 } 73 74 return result 75 } 76 }