Amazon
  • Introduction
  • Phone Interview I
    • 53.Reverse Words in a String
    • 31.Partition Array
  • Phone Interview II
    • 167.Add Two Numbers
    • 88.Lowest Common Ancestor
  • Onsite I
    • 655.Big Integer Addition
    • 221.Add Two Numbers II
  • Onsite II
    • 158.Two Strings Are Anagrams
    • 171.Anagrams
    • 386.Longest Substring with At Most K Distinct Characters
  • Onsite III
    • 479.Second Max of Array
    • 589.Connecting Graph
  • Onsite IV
    • 532.Reverse Pairs
  • 2022
    • OA
      • work simulation
      • Greyness
      • NearestRetailer
      • Sum of Scores of Subarray
      • StrengthOfPassword
      • ProductOf1
      • Move 0/1 InArray
      • Max deviation among all substrings
      • AWS power consumption
      • searchWordResultWord
      • maxOperationOfString
      • MinHealthGame
      • EarliestMonth
      • Package ship
      • RatingConsectiveDecresing
      • LinkedListSum
      • MovingBoxes
      • ValidString
      • MaxValueAfterRemovingFromString
      • Subtree with Maximum Average
    • VO
      • 2022.3
    • BQ
      • doc
      • 2022.4
      • Freq Question
      • 11大类BQ和Follow-ups
      • Page 1
      • BQ 100
      • 35 behavioral questions asked in 95% of Amazon interviews with examples
      • Page 2
      • 反向BQ
    • LP
      • LP-1
      • LP-2
    • SD
      • Design Amazon Prime Video Recommendation System
      • Amazon Order system
    • OOD
      • Linux Find Command
      • Amazon Locker
    • AWS Identity call
    • Interviews
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  1. Onsite III

479.Second Max of Array

1.Description(Easy)

Find the second max number in a given array.

Notice

You can assume the array contains at least two numbers.

Example

Given[1, 3, 2, 4], return3.

Given[1, 2], return1.

2.Code

这个题不能用排序然后选择第二小的来做,因为当所有的值都一样大时应该输出这个值,而排序的话就会报错。

最开始的两个值大小,得出first max和second max,在后面的每个值和这两个值比大小来更新first 和second.

public int secondMax(int[] nums) {
        int first=Math.max(nums[0], nums[1]);
        int second=Math.min(nums[0], nums[1]);

        for(int i=2;i<nums.length;i++){
            if(nums[i]>first){
                second=first;
                first=nums[i];               
            }
            else if(nums[i]>second){
                second=nums[i];
            }
        }
        return second;
    }
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Last updated 5 years ago

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