}elseif(sz ==2){returnmax(nums[0], nums[1]); }inta = nums[0];intb =0;intc =0;boolsecond =false;// the second biggest number has foundboolthird =false;// the third biggest number has foundfor(inti =1; i < sz; i++){// cout<<a<<", "<<b<<", "<<c<<endl;if(a =...
Explanation: Note that the third maximum here means the third maximum distinct number. Both numbers with value 2 are both considered as second maximum. 找出数组中数组第三大的元素,如果不存在这个元素则返回数组中值最大的元素。 首先对数组进行去重,使用STL中的unique函数,然后判断数组的大小,如果存在第三...
LeetCode 414. Third Maximum Number 简介:给定一个非空数组,返回此数组中第三大的数。如果不存在,则返回数组中最大的数。要求算法时间复杂度必须是O(n)。 Description Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum numbe...
nums.remove(max(nums)) return max(nums) def thirdMax2(self, nums): v = [float('-inf'), float('-inf'), float('-inf')] for num in nums: if num not in v: if num > v[0]: v = [num, v[0], v[1]] print v elif num > v[1]: v = [v[0], num, v[1]] print v...
leetcode 485[easy]---Max Consecutive Ones Givenabinaryarray, findthemaximumnumberofconsecutive 1sinthisarray.Example 1: Note:Theinputarraywill only contain 0 and 1.Thelengthofinputarrayisapositive integer and willnot 作业(十二)——414. Third Maximum Number ...
leetcode/problems/src/array/ThirdMaximumNumber.java/ Jump to 64 lines (55 sloc)1.65 KB RawBlame packagearray; /** * Created by gouthamvidyapradhan on 25/03/2017. Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximu...
Explanation: Note that the third maximum here means the third maximum distinct number. Both numbers with value 2 are both considered as second maximum. 来源:力扣(LeetCode) 分析:题目大意就是给你一个不为空的数组(并非有序或无序),让你找出数组中第三大的元素值,如果找不到就返回其最大值,题意就...
Input:[2,2,3,1]Output:1Explanation:Note that the third maximum here means the third maximum distinct number.Both numberswithvalue2are both consideredassecond maximum. 我的答案: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution{publicintthirdMax(int[]nums){sort(nums);int a=1;in...
int thirdMax(vector<int>& nums) { set<int> num; for(int i =0;i<nums.size();i++) num.insert(nums[i]); set<int>::iterator itr = --num.end(); for(int j =num.size()-1;j>=0;j--,itr--) if(j==num.size()-3) ...
Explanation: Note that the third maximum here means the third maximum distinct number. Both numbers with value 2 are both considered as second 1. 2. 3. 4. 5. 6. public class Solution { public int thirdMax(int[] nums) { Integer max1 = null; ...