zoukankan      html  css  js  c++  java
  • [Swift]LeetCode496. 下一个更大元素 I | Next Greater Element I

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
    ➤微信公众号:山青咏芝(shanqingyongzhi)
    ➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
    ➤GitHub地址:https://github.com/strengthen/LeetCode
    ➤原文地址:https://www.cnblogs.com/strengthen/p/9805282.html 
    ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
    ➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

    The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

    Example 1:

    Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
    Output: [-1,3,-1]
    Explanation:
        For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
        For number 1 in the first array, the next greater number for it in the second array is 3.
        For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

    Example 2:

    Input: nums1 = [2,4], nums2 = [1,2,3,4].
    Output: [3,-1]
    Explanation:
        For number 2 in the first array, the next greater number for it in the second array is 3.
        For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

    Note:

    1. All elements in nums1 and nums2 are unique.
    2. The length of both nums1 and nums2 would not exceed 1000.

     给定两个没有重复元素的数组 nums1 和 nums2 ,其中nums1 是 nums2 的子集。找到 nums1 中每个元素在 nums2 中的下一个比其大的值。

    nums1 中数字 x 的下一个更大元素是指 x 在 nums2 中对应位置的右边的第一个比 x 大的元素。如果不存在,对应位置输出-1。

    示例 1:

    输入: nums1 = [4,1,2], nums2 = [1,3,4,2].
    输出: [-1,3,-1]
    解释:
        对于num1中的数字4,你无法在第二个数组中找到下一个更大的数字,因此输出 -1。
        对于num1中的数字1,第二个数组中数字1右边的下一个较大数字是 3。
        对于num1中的数字2,第二个数组中没有下一个更大的数字,因此输出 -1。

    示例 2:

    输入: nums1 = [2,4], nums2 = [1,2,3,4].
    输出: [3,-1]
    解释:
        对于num1中的数字2,第二个数组中的下一个较大数字是3。
        对于num1中的数字4,第二个数组中没有下一个更大的数字,因此输出 -1。
    

    注意:

    1. nums1nums2中所有元素是唯一的。
    2. nums1nums2 的数组大小都不超过1000。

    16ms

     1 class Solution {
     2     func nextGreaterElement(_ findNums: [Int], _ nums: [Int]) -> [Int] {
     3         
     4         // 9,8,7,6,5,10 -> 10 is the next greater for all
     5         var result = [Int]()
     6         var dict = [Int: Int]()
     7         var stack = [Int]()
     8         
     9         for num in nums {
    10             while !stack.isEmpty && stack.last! < num {
    11                 let last = stack.removeLast()
    12                 dict[last] = num
    13             }
    14             
    15             stack.append(num)
    16         }
    17         
    18         for num in findNums {
    19             result.append(dict[num] ?? -1)
    20         }
    21         
    22         return result
    23     }
    24 }

    20ms

     1 class Solution {
     2     func nextGreaterElement(_ findNums: [Int], _ nums: [Int]) -> [Int] {
     3         
     4         var ans = [Int]()
     5         var stack = [Int]()
     6         var dict = [Int:Int]()
     7         
     8         for (index, item) in nums.enumerated() {
     9             dict[item] = index
    10             ans.append(0)
    11         }
    12         
    13         for val in nums {
    14             while stack.count > 0 && val > stack.last! {
    15                 ans[dict[stack.last!]!] = val
    16                 stack.removeLast()
    17             }
    18             stack.append(val)
    19         }
    20         var newAns = [Int]()
    21         for val in findNums {
    22             if let x = dict[val] {
    23                 newAns.append(ans[x]==0 ? -1 : ans[x])
    24             }
    25         }
    26         return newAns
    27     }
    28 }

    24ms

     1 class Solution {
     2     func nextGreaterElement(_ findNums: [Int], _ nums: [Int]) -> [Int] {
     3         var dict = [Int:Int]()
     4         var stack = [Int]()
     5     
     6         for num in nums {
     7             while stack.count > 0 && stack.last! < num {
     8                 dict[stack.popLast()!] = num
     9             }
    10             stack.append(num)
    11         }
    12 
    13         return findNums.map { dict[$0] ?? -1 }
    14     }
    15 }

    32ms

     1 class Solution {
     2     func findindex(_ nums: [Int], _ search: Int, _ count: Int) -> Int {
     3         for i in 0..<count {
     4             if (nums[i] == search) {
     5                 return i
     6             }
     7         }
     8         return -1
     9     }
    10     func nextGreaterElement(_ findNums: [Int], _ nums: [Int]) -> [Int] {
    11         var output = [Int]()
    12         var index = -1
    13         var flag = false
    14         let count = nums.count
    15         for i in findNums {
    16             index = findindex(nums, i, count)
    17             if (index != (count-1)) {
    18                 flag = false
    19                 for j in nums[index..<count] {
    20                     if (i<j) {
    21                         output.append(j)
    22                         flag = true
    23                         break
    24                     }
    25                 }
    26                 if (!flag) {
    27                     output.append(-1)
    28                 }
    29             } else {
    30                 output.append(-1)
    31             }
    32         }
    33         return output
    34     }
    35 }

    40ms

     1 class Solution {
     2     func nextGreaterElement(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
     3         var dict = [Int: Int]()
     4     
     5         for (index, num) in nums2.enumerated() {
     6             dict[num] = index
     7         }
     8 
     9         var result: [Int] = []
    10         for num in nums1 {
    11             var index: Int = dict[num]!
    12             var max: Int = -1
    13             while index < nums2.count {
    14                 if num < nums2[index] {
    15                     max = nums2[index]
    16                     break
    17                 }
    18 
    19                 index += 1
    20             }
    21 
    22             result.append(max)
    23         }
    24 
    25         return result
    26     }
    27 }

    44ms

     1 class Solution {
     2     func nextGreaterElement(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
     3         var nextGreater = [Int]()
     4         var stack = [Int]()
     5         var dict = [Int:Int]()
     6 
     7         for num in nums2 {
     8             while !stack.isEmpty && num > stack.last! {
     9                 dict[stack.removeLast()] = num
    10             }
    11 
    12             stack.append(num)
    13         }
    14 
    15         for num in nums1 {
    16             nextGreater.append(dict[num] ?? -1)
    17         }
    18 
    19         return nextGreater
    20     }
    21 }

    68ms

     1 class Solution {
     2     func nextGreaterElement(_ findNums: [Int], _ nums: [Int]) -> [Int] {
     3             var result: [Int] = []
     4             var item = findNums.count-1
     5             while item >= 0 {
     6                 var itemResult = -1
     7                 var itemFinded = -1
     8                 for index in 0..<nums.count {
     9                     if findNums[item] == nums[index] {
    10                         itemFinded = index
    11                     }
    12                     if findNums[item] < nums[index] && itemFinded != -1 {
    13                         itemResult = nums[index]
    14                         break
    15                     }
    16                 }
    17                 result.insert(itemResult, at: 0)
    18                 item = item - 1
    19             }
    20             return result
    21         }
    22 }

    72ms

     1 class Solution {
     2     func nextGreaterElement(_ findNums: [Int], _ nums: [Int]) -> [Int] {
     3          var res:[Int] = []
     4         for i in 0..<findNums.count {
     5             for j in 0..<nums.count{
     6                 if findNums[i] == nums[j] {
     7                     for k in j...nums.count-1 {
     8                         if (findNums[i] < nums[k]) {
     9                             res.append(nums[k])
    10                             break
    11                         }else if (k == (nums.count-1)){
    12                             res.append(-1)
    13                         }
    14                     }
    15                     
    16                 }
    17             }
    18         }
    19         return res
    20     }
    21 }

    72ms

     1 class Solution {
     2     func nextGreaterElement(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
     3         var nextGreater = [Int]()
     4         var stack = [Int]()
     5         var dict = [Int:Int]()
     6 
     7         for num in nums2 {
     8             while !stack.isEmpty && num > stack.last! {
     9                 dict[stack.removeLast()] = num
    10             }
    11 
    12             stack.append(num)
    13         }
    14 
    15         for num in nums1 {
    16             nextGreater.append(dict[num] ?? -1)
    17         }
    18 
    19         return nextGreater
    20     }
    21 }
  • 相关阅读:
    通过java代码获取jvm信息和系统信息
    java cp与java jar的区别
    vue下实现WebRTC
    MANIFEST.MF文件详解
    element 前端排序 与 后端排序
    JAVA获取CPUID、主板序列号、硬盘序列号、MAC地址(自己验证过)
    PHP常用代码大全
    程序员从初级到中级10个秘诀
    移动平台还有哪些创业机会
    程序员招聘:如何识别真正的程序员
  • 原文地址:https://www.cnblogs.com/strengthen/p/9805282.html
Copyright © 2011-2022 走看看