LeetCode 414. Third Maximum Number (Java)

题目：

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: 1
Explanation: The third maximum is 1.

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

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

解答：

//题目中要求时间复杂度为O(n)
//使用3次遍历数组，分别找出第一大，第2大以及第三大的数
class Solution {
    public int thirdMax(int[] nums) {
        int Maxone=Integer.MIN_VALUE;
        int Maxsecond=Integer.MIN_VALUE;
        int Maxthird=Integer.MIN_VALUE;
        int flag=0;
        if(nums.length==1){
            return nums[0];
        }
        if(nums.length==2){
            return Math.max(nums[0],nums[1]);
        }
        for(int i=0;i<nums.length;i++){
            if(nums[i]>Maxone){
                Maxone=nums[i];
            }
        }
        for(int i=0;i<nums.length;i++){
            if(nums[i]>Maxsecond&&nums[i]<Maxone){
                Maxsecond=nums[i];
            }
        }
        for(int i=0;i<nums.length;i++){
            if(nums[i]>=Maxthird&&nums[i]<Maxsecond){
            //注意flag标志位的设立
                Maxthird=nums[i];
                flag=1;
            }
        }
        if(flag==1){
            return Maxthird;
        } 
        return Maxone;
    }
}

第一次提交中没有设立flag标志位，导致提交错误。

第二次提交中，第三次遍历数组求第三大值时，判断条件为if(nums[i]>Maxthird&&nums[i]而不是if(nums[i]>=Maxthird&&nums[i]，导致提交错误