Third Maximum Number

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414.Third Maximum Number

Given an integer array nums, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.

Example 1:

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Input: nums = [3,2,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.

Example 2:

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

Example 3:

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Input: nums = [2,2,3,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2's are counted together since they have the same value).
The third distinct maximum is 1.

排序：

时间复杂度：O(nlogn)

空间复杂度：O(logn)

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class Solution {
public:
    int thirdMax(vector<int> &nums) {
        sort(nums.begin(), nums.end(), greater<>());
        for (int i = 1, diff = 1; i < nums.size(); ++i) {
            if (nums[i] != nums[i - 1] && ++diff == 3) {
                return nums[i];
            }
        }
        return nums[0];
    }
};

有序集合：

时间复杂度：O(n)

空间复杂度：O(1)

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class Solution {
public:
    int thirdMax(vector<int> &nums) {
        set<int> s;
        for (int num : nums) {
            s.insert(num);
            if (s.size() > 3) {
                s.erase(s.begin());
            }
        }
        return s.size() == 3 ? *s.begin() : *s.rbegin();
    }
};

一次遍历：

时间复杂度：O(n)

空间复杂度：O(1)

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class Solution {
public:
    int thirdMax(vector<int> &nums) {
        long a = LONG_MIN, b = LONG_MIN, c = LONG_MIN;
        for (long num : nums) {
            if (num > a) {
                c = b;
                b = a;
                a = num;
            } else if (a > num && num > b) {
                c = b;
                b = num;
            } else if (b > num && num > c) {
                c = num;
            }
        }
        return c == LONG_MIN ? a : c;
    }
};
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class Solution {
public:
    int thirdMax(vector<int> &nums) {
        int *a = nullptr, *b = nullptr, *c = nullptr;
        for (int &num : nums) {
            if (a == nullptr || num > *a) {
                c = b;
                b = a;
                a = &num;
            } else if (*a > num && (b == nullptr || num > *b)) {
                c = b;
                b = &num;
            } else if (b != nullptr && *b > num && (c == nullptr || num > *c)) {
                c = &num;
            }
        }
        return c == nullptr ? *a : *c;
    }
};