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  • [Swift]LeetCode1 .两数之和 | Two Sum

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    ➤原文地址:https://www.cnblogs.com/strengthen/p/9697856.html 
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    Given an array of integers, return indices of the two numbers such that they add up to a specific target.

    You may assume that each input would have exactly one solution, and you may not use the same element twice.

    Example:

    Given nums = [2, 7, 11, 15], target = 9,
    Because nums[0] + nums[1] = 2 + 7 = 9,
    return [0, 1].

    给定一个整数数组和一个目标值,找出数组中和为目标值的两个数。你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用。

    示例:

    给定 nums = [2, 7, 11, 15], target = 9
    因为 nums[0] + nums[1] = 2 + 7 = 9
    所以返回 [0, 1]

    一遍哈希表

    我们可以一次完成。在进行迭代并将元素插入到表中的同时,我们还会回过头来检查表中是否已经存在当前元素所对应的目标元素。如果它存在,那我们已经找到了对应解,并立即将其返回。

    Runtime: 32 ms
    Memory Usage: 19.3 MB
     1 class Solution {
     2     func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
     3         var table:[Int:Int] = [Int:Int]()        
     4         for (firstIndex, num) in nums.enumerated() {
     5             var res: Int = target - num
     6             //可选链接
     7             if let secondIndex = table[res] 
     8             {
     9                 return [secondIndex,firstIndex]
    10             } 
    11             else 
    12             {
    13                 table[num] = firstIndex
    14             }
    15         }
    16         return [-1,-1]
    17     }
    18 }

    36ms 
     1 class Solution {
     2     func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
     3         var out: [Int] = []
     4        
     5         var dict:[Int: Int] = [:]     
     6         for (index,num) in nums.enumerated() {
     7             if let sum = dict[target - num] {
     8                 out.append(index)
     9                 out.append(sum)
    10             } 
    11             dict[num] = index
    12         }
    13         return out
    14     }
    15 }

     1 class Solution {
     2     func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
     3         for i in 0...nums.count-1 {
     4             if i == nums.count-1 { break; }
     5             for j in i+1...nums.count-1 {
     6                 let sum = nums[i] + nums[j]
     7                 if sum == target {
     8                     return [i, j]
     9                 }
    10             }            
    11         }
    12         return [Int]()
    13     }
    14 }

     1 class Solution {
     2     typealias ArrayIndex = (array: [Int], startIndex: Int)
     3     
     4     func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
     5         var arrayTest = [ArrayIndex]()
     6         arrayTest.append(ArrayIndex(nums, 0))
     7         
     8         while arrayTest.count != 0 {
     9             let values = arrayTest.remove(at: 0)
    10             let splitValues = splitArray(nums: values)
    11             if let result = scanTwoArray(nums1: splitValues.nums1, nums2: splitValues.nums2, target: target) {
    12                 return result
    13             } else {
    14                 arrayTest.append(splitValues.nums1)
    15                 arrayTest.append(splitValues.nums2)
    16             }
    17         }
    18         return [0,1]
    19     }
    20     
    21     func splitArray(nums: ArrayIndex) -> (nums1: ArrayIndex, nums2: ArrayIndex) {
    22         let ct = nums.array.count
    23         let half = ct / 2
    24         let leftSplit = nums.array[0 ..< half]
    25         let rightSplit = nums.array[half ..< ct]
    26         return ((Array(leftSplit), nums.startIndex), (Array(rightSplit), nums.startIndex + half))
    27     }
    28     
    29     func scanTwoArray(nums1: ArrayIndex, nums2: ArrayIndex, target: Int) -> [Int]? {
    30         for i in 0..<nums1.array.count {
    31             for j in 0..<nums2.array.count {
    32                 if nums1.array[i] + nums2.array[j] == target {
    33                     return [i + nums1.startIndex, j + nums2.startIndex]
    34                 }
    35             }
    36         }
    37         
    38         return nil
    39     }
    40 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/9697856.html
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