Leetcode——Third Maximum Number

Question

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.

Solution 1

用两个标识符表示，第二大的数和第三大的数是否已经赋值。

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        if (nums.size() == 1)
            return nums[0];
        if (nums.size() == 2)
            return max(nums[0], nums[1]);

        long one = nums[0];
        long two = LONG_MIN;
        long three = LONG_MIN;
        bool two_flag = true;
        bool three_flag = true;

        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] > one) {
                if (one > two) {
                    if (two > three) {
                        three = two;
                        three_flag = false;
                    }
                    two = one;
                    two_flag = false;
                }
                one = nums[i];
            }
            else if ((two_flag || nums[i] > two) && nums[i] != one) {
                if (two > three) {
                    three = two;
                    three_flag = false;
                }
                two = nums[i];
                two_flag = false;
            }
            else if ((three_flag || nums[i] > three) && nums[i] != one && nums[i] != two) {
                three = nums[i];
                three_flag = false;
            }

        }
        if (!three_flag)
            return three;
        else
            return one;
    }
};

Solution 2

用STL set,因为set中最多有3个元素，因此操作都是常量时间的，这样就可以保证总的时间是O(n)，而且set会自动排序，底层实现是二叉排序树，因为有利于比较。

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        set<int> s;
        for (int i = 0; i < nums.size(); i++) {
            if (s.size() < 3)
                s.insert(nums[i]);
            else {
                set<int>::iterator it = s.begin();
                if (s.find(nums[i]) == s.end() && (nums[i] > *(it) || nums[i] > *(++it) || nums[i] > *(++it)) ) {
                    s.erase(it);
                    s.insert(nums[i]);
                }
            }
        }
        if (s.size() == 3)
            return *(s.begin());
        else
            return *(s.rbegin());
    }
};