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)。
示例 1:
输入: [3, 2, 1]
输出: 1
解释: 第三大的数是 1.
示例 2:
输入: [1, 2]
输出: 2
解释: 第三大的数不存在, 所以返回最大的数 2 .
示例 3:
输入: [2, 2, 3, 1]
输出: 1
解释: 注意,要求返回第三大的数,是指第三大且唯一出现的数。
存在两个值为2的数,它们都排第二。
详见:https://leetcode.com/problems/third-maximum-number/description/
Java实现:
class Solution {
public int thirdMax(int[] nums) {
long first=Long.MIN_VALUE;
long second=Long.MIN_VALUE;
long third=Long.MIN_VALUE;
for(int num:nums){
if(first<num){
third=second;
second=first;
first=num;
}else if(num>second&&num<first){
third=second;
second=num;
}else if(num>third&&num<second){
third=num;
}
}
return (third==Long.MIN_VALUE||third==second)?(int)first:(int)third;
}
}
C++实现:
class Solution {
public:
int thirdMax(vector<int>& nums) {
long first = LONG_MIN, second = LONG_MIN, third = LONG_MIN;
for (int num : nums)
{
if (num > first)
{
third = second;
second = first;
first = num;
}
else if (num > second && num < first)
{
third = second;
second = num;
}
else if (num > third && num < second)
{
third = num;
}
}
return (third == LONG_MIN || third == second) ? first : third;
}
};
参考:https://www.cnblogs.com/grandyang/p/5983113.html