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  • leetcode 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: 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.

    求一堆数的前3大,一般排序就可以了,但是题目要求O(n)的时间去做

    本题目也是6, 

    3 2 2 2 2 2 第三大是 3

    2 2 2 第三大是2

    没关系,这些题目中都已经说明了

    class Solution {
    public:
        int thirdMax(vector<int>& nums) {
            int n = nums.size();
            long Max = LONG_MIN;
            if (n < 3) {
                for (int i = 0; i < n; ++i) if (nums[i] > Max) Max = nums[i];
                return Max;
            }
            else {
                long Max2 = LONG_MIN;
                long Max3 = LONG_MIN;
                for (int i = 0; i < n; ++i) {
                    if (nums[i] > Max) Max3 = Max2 ,Max2 = Max, Max = nums[i];
                    else if (nums[i] > Max2 && nums[i] < Max) Max3 = Max2,Max2 = nums[i];
                    else if (nums[i] > Max3 && nums[i] < Max2) Max3 = nums[i];
                }
                if (Max3 == LONG_MIN) Max3 = Max;
                return Max3;
            }
        }
    };
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  • 原文地址:https://www.cnblogs.com/pk28/p/7257160.html
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