[LeetCode]414. Third Maximum Number 解题报告(C++)
题目描述
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.
题目大意
- 给定一个非空的数组.
- 返回第三大的数字.
- 注意.不存在返回第一大.
- 要求时间复杂度O(n)
- 且第三大和第二大不能相同.严格小于!
解题思路
方法1:
- 遍历数组找最大.第二大.第三大.
- 注意坑!!! 初始化要用 LONG_MIN
代码实现:
class Solution0 {
public:
int thirdMax(vector<int>& nums) {
long mx1 = LONG_MIN, mx2 = LONG_MIN, mx3 = LONG_MIN;
for (int x : nums) {
if (x > mx1) {
mx3 = mx2;
mx2 = mx1;
mx1 = x;
}
else if (x<mx1&&x>mx2) {
mx3 = mx2;
mx2 = x;
}
else if (x > mx3&&x < mx2) {
mx3 = x;
}
}
return(mx3 == LONG_MIN || mx3 ==mx2 ? mx1 : mx3);
}
};
方法2:
- 利用
set
的自动排序和自动去重功能
代码实现:
class Solution {
public:
int thirdMax(vector<int>& nums) {
set<int> s;
for (auto x : nums) {
s.insert(x);
if (s.size() > 3) {
s.erase(s.begin());
}
}
return s.size() == 3 ? *s.begin() : *s.rbegin();
}
};
小结
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