414. Third Maximum Number

Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:

Input: [3, 2, 1]Output: 1Explanation: The third maximum is 1.

Example 2:

Input: [1, 2]Output: 2Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: [2, 2, 3, 1]Output: 1Explanation: Note that the third maximum here means the third maximum distinct number.Both numbers with value 2 are both considered as second maximum.

法1：使用快排中的“切分” O(NlgN)

public class Solution {    public int thirdMax(int[] nums){        HashSet<Integer> hashset=new HashSet<>(nums.length);        for(int n : nums)        hashset.add(n);                int[] arr=new int[hashset.size()];        int cnt=0;        for(int n :hashset)        arr[cnt++]=n;                int len=arr.length;                if(len<3)        {        int max=Integer.MIN_VALUE;        for(int i=0;i<len;i++)        max=Math.max(arr[i], max);        return max;        }                              return findkth(arr, len-3, 0, len-1);            }public int findkth(int[] nums,int k,int lo,int hi){int kth=partition(nums, lo, hi);int ele=nums[kth];if(kth>k)return findkth(nums, k, lo, kth);else if(kth<k)return findkth(nums, k, kth+1, hi);else {return ele;}}public int partition(int[] nums,int lo,int hi){if(lo==hi)return lo;int num=nums[lo];int i=lo,j=hi+1;while(true){while(nums[++i]<num)if(i==hi)break;while(nums[--j]>num)if(j==lo)break;if(i>=j)break;swap(nums, i, j);}swap(nums, lo, j);return j;}public static void swap(int[] arr ,int i, int j){int temp=arr[i];arr[i]=arr[j];arr[j]=temp;}}

法2 ，使用三个变量直接检索，摘自

https://discuss.leetcode.com/topic/62236/java-solution-in-0ms-run-time-o-n-and-space-o-1/15

public int thirdMax(int[] nums) {    long max = Long.MIN_VALUE, mid = max, min = max;            for (int ele : nums) {        if (ele > max) {            min = mid;            mid = max;            max = ele;        } else if (max > ele && ele > mid) {            min = mid;            mid = ele;        } else if (mid > ele && ele > min) {            min = ele;        }    }            return (int)(min != Long.MIN_VALUE ? min : max);}