Leetcode 414.第三大的数

原题链接:Leetcode 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:

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:

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:

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.

Constraints:

  • 1 <= nums.length <= 104
  • -231 <= numsi <= 231 - 1

Follow up: Can you find an O(n) solution?


方法一:遍历数组

思路:

题目要求返回第三大的数,以及当没有三个不同数的情况下返回第二个数

那么就维护三个值:first、second、third,最后分情况返回即可

比较麻烦的就是初始值,需要设置一个最小的正数,这里用 LONG_LONG_MIN

C++ 代码:

cpp 复制代码
class Solution {
public:
    // const long long Min = -(1LL << 31) - 1;
    int thirdMax(vector<int>& nums) {
        long first = LONG_LONG_MIN, second = LONG_LONG_MIN, third = LONG_LONG_MIN;    // 当前最大的三个数
        
        // 遍历的同时维护这三个数
        for(int i = 0; i < nums.size(); i++ ){
            if(nums[i] > first){
                third = second;
                second = first;
                first = nums[i];
            }
            else if(nums[i] < first && nums[i] > second){
                third = second;
                second = nums[i];
            }
            else if(nums[i] < second && nums[i] > third){
                third = nums[i];
            }
        }

        if(third == LONG_LONG_MIN)
            return first;
        return third;
    }
};
相关推荐
hanlin037 小时前
刷题笔记:力扣第32题-最长有效括号
笔记·算法·leetcode
All for pursuit.7 小时前
【动态规划-21】312.戳气球
数据结构·c++·算法·leetcode·动态规划
碧海银沙音频科技研究院9 小时前
基于炬芯ATS3089X自研AEC声学回声消除算法
算法
通问AI10 小时前
在两台 Mac Studio 上跑 800B 模型:MLX + RDMA over Thunderbolt 实测与四道门选型法
算法
ZywOo62310 小时前
《算法竞赛进阶指南》0x03 前缀和与差分
算法
Scintilla_y10 小时前
DAY2 滑动窗口
算法·leetcode·散列表
lvwangshu10 小时前
笛卡尔树
算法·图论
026d2310 小时前
状态信号只知“其然”不知“其所以然”:变工况机械故障诊断的物理盲区
算法
姜鱼问生10 小时前
备份和服务器在同一台机器上,等于没备份
算法·架构
姜鱼问生11 小时前
别再 chmod 777 了,用 ACL 给它开一扇窗
后端·算法·架构