https://leetcode-cn.com/problems/third-maximum-number/
/*
* @Author: Goog Tech
* @Date: 2020-07-25 10:45:27
* @Description: https://leetcode-cn.com/problems/third-maximum-number/
* @FilePath: \leetcode-googtech\#414. Third Maximum Number\Solution.java
*/
class Solution {
// Long.MIN_VALUE=-9223372036854775808(-2的63次方)
private long MIN_VALUE = Long.MIN_VALUE;
public int thirdMax(int[] nums) {
// 初始化第一,第二以及第三大元素
long firstElement = MIN_VALUE,secondElement = MIN_VALUE,thirdElement = MIN_VALUE;
// 遍历数组,获取第一,第二以及第三大元素
for(int num : nums) {
if(num > firstElement) {
thirdElement = secondElement;
secondElement = firstElement;
firstElement = num;
}else if(secondElement < num && num < firstElement) {
thirdElement = secondElement;
secondElement = num;
}else if (thirdElement < num && num < secondElement) {
thirdElement = num;
}
}
// 若无第三大元素则返回数组中的最大值
return thirdElement==MIN_VALUE ? (int)firstElement : (int)thirdElement;
}
}
'''
@Author: Goog Tech
@Date: 2020-07-25 10:45:35
@Description: https://leetcode-cn.com/problems/third-maximum-number/
@FilePath: \leetcode-googtech\#414. Third Maximum Number\Solution.py
'''
class Solution(object):
def thirdMax(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
# 初始化第一,第二以及第三大元素
# -sys.maxsize = -9223372036854775807
first = second = thrid = -sys.maxsize
# 遍历数组,获取第一,第二以及第三大元素的值
for num in nums:
if num > first:
thrid, second, first = second, first, num
elif second < num < first:
thrid, second = second, num
elif thrid < num < second:
thrid = num
# 若无第三大元素则返回最大元素的值
return first if thrid == -sys.maxsize else thrid