LeetCode: Third Maximum Number

最新推荐文章于 2021-07-30 09:59:34 发布

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最新推荐文章于 2021-07-30 09:59:34 发布

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分类专栏： LeetCode 文章标签： LeetCode array maximum

本文链接：https://blog.youkuaiyun.com/u012309913/article/details/54986968

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本文介绍了一个算法问题：如何在一个整数数组中找到第三大的数。如果不存在第三大的数，则返回最大的数。文章提供了C++代码实现，展示了如何通过一次遍历来解决此问题，并确保了时间复杂度为O(n)。

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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.

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        if (nums.size() == 1) {
            return nums[0];
        }
        else if (nums.size() == 2) {
            if (nums[0] > nums[1]) {
                return nums[0];
            }
            else return nums[1];
        }
        
        vector<int> left;
        int max1 = -2147483648;
        int max2 = -2147483648;
        int max3 = -2147483648;
        int n = nums.size();
        bool appeared = false;
        
        for (int i = 0; i < n; ++i)
        {
            if (nums[i] == -2147483648)
            {
                appeared = true;
            }
            if (nums[i] >= max1)
            {
                if (nums[i] > max1) {
                max3 = max2;
                max2 = max1;
                max1 = nums[i];
                }
            }
            else if (nums[i] >= max2)
            {
                if (nums[i] > max2) {
                
                max3 = max2;
                max2 = nums[i];
                }
            }
            else if (nums[i] >= max3)
            {
                max3 = nums[i];
            }
        }
        
        if (!appeared)
        {
            if (max3 == -2147483648) {
                return max1;
            }
            else return max3;
        }
        else if (max2 > max3) {
            return max3;
        }
        else return max1;
    }
};