LeetCode-414. Third Maximum Number

最新推荐文章于 2023-12-26 12:49:54 发布

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最新推荐文章于 2023-12-26 12:49:54 发布

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本文介绍了一种算法，用于从非空整数数组中找出第三大的数。若不存在第三大的数，则返回最大的数。该算法首先对数组进行排序，然后通过计数器找到第三个最大元素，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.

先排序，然后设置计数器找到第三个大的元素。注意如果找不到，那么就返回最大的那个元素。

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