leetcode 414. Third Maximum Number

标签：value   说明   com   return   complex   ret   排序   empty   pre   

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.

求一堆数的前3大，一般排序就可以了，但是题目要求O(n)的时间去做

本题目也是6, 

3 2 2 2 2 2 第三大是 3

2 2 2 第三大是2

没关系，这些题目中都已经说明了

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        int n = nums.size();
        long Max = LONG_MIN;
        if (n < 3) {
            for (int i = 0; i < n; ++i) if (nums[i] > Max) Max = nums[i];
            return Max;
        }
        else {
            long Max2 = LONG_MIN;
            long Max3 = LONG_MIN;
            for (int i = 0; i < n; ++i) {
                if (nums[i] > Max) Max3 = Max2 ,Max2 = Max, Max = nums[i];
                else if (nums[i] > Max2 && nums[i] < Max) Max3 = Max2,Max2 = nums[i];
                else if (nums[i] > Max3 && nums[i] < Max2) Max3 = nums[i];
            }
            if (Max3 == LONG_MIN) Max3 = Max;
            return Max3;
        }
    }
};

leetcode 414. Third Maximum Number

标签：value   说明   com   return   complex   ret   排序   empty   pre   

原文地址：http://www.cnblogs.com/pk28/p/7257160.html