Description
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.
分析
题目的意思是:
- 用三个变量A[0], A[1], A[2]来分别保存第一大,第二大,和第三大的数,然后遍历数组,如果遍历到的数字大于当前第一大的数A[0],那么三个变量各自错位赋值,如果当前数字大于A[1],小于A[0],那么就更新A[1]和A[2],如果当前数字大于A[2],小于A[1],那就只更新A[0]。
- 注意这里有个坑,就是初始化要用长整型long的最小值,否则当数组中有INT_MIN存在时,程序就不知道该返回INT_MIN还是最大值first了。
代码
class Solution {
public:
int thirdMax(vector<int>& nums) {
vector<long> A(3,LONG_MIN);
for(int i=0;i<nums.size();i++){
if(A[0]<nums[i]){
A[2]=A[1];
A[1]=A[0];
A[0]=nums[i];
}else if(A[1]<nums[i]&&nums[i]<A[0]){
A[2]=A[1];
A[1]=nums[i];
}else if(A[2]<nums[i]&&nums[i]<A[1]){
A[2]=nums[i];
}
}
return (A[2]==LONG_MIN||A[2]==A[1]) ? A[0]:A[2];
}
};
参考文献
[LeetCode] Third Maximum Number 第三大的数