LIntCode 1250: Third Maximum Number_1944. 被同一个人至少使用过三次的共享单车 lintcode-优快云博客

本文介绍了一种高效算法，用于从非空整数数组中找出第三大的数。如果不存在，则返回最大的数。通过实例展示了如何使用O(n)的时间复杂度解决此问题。

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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
Example 1:

Input: num = [3, 2, 1]
Output: 1
Explanation: The third maximum is 1.
Example 2:

Input: num = [1, 2]
Output: 2
Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
Example 3:

Input: num = [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.

解法1：
注意:
1）top0，top1，top2都初始化为INT_MIN。
2）迭代顺序为top2=top1, top1=top0, top0=nums[i]。反过来那3个数都变成一样的，就错了。
3）注意nums[i]<top0和nums[i]<top1这两个条件。没有的话那相同的数也会造成top2/top1变化。就错了。

class Solution {
public:
    /**
     * @param nums: the array
     * @return: the third maximum number in this array
     */
    int thirdMax(vector<int> &nums) {
        int n = nums.size();

        int top0, top1, top2;
        
        top0 = INT_MIN, top1 = INT_MIN, top2 = INT_MIN;
        for (int i = 0; i < n; ++i) {
            if (nums[i] > top0) {
                top2 = top1;
                top1 = top0;
                top0 = nums[i];
            } else if (nums[i] > top1 && nums[i] < top0) {
                top2 = top1;
                top1 = nums[i];
            } else if (nums[i] > top2 && nums[i] < top1) {
                top2 = nums[i];
            }
        }
        //cout<<"top0="<<top0<<" top1="<<top1<<" top2="<<top2<<endl;
        if (n < 3) return top0;
        if (top2 == top1) return top0;
        return top2;
    }
};