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.
题解如下:
class Solution {
public int thirdMax(int[] nums) {
Integer max1 = null;
Integer max2 = null;
Integer max3 = null;
for(Integer num:nums) {
if(num.equals(max1) || num.equals(max2) || num.equals(max3))
continue;
else if(max1 == null || num > max1) {
max3 = max2;
max2 = max1;
max1 = num;
} else if(max2 == null || num > max2) {
max3 = max2;
max2 = num;
} else if(max3 == null || num > max3) {
max3 = num;
}
}
return max3 == null ? max1:max3;
}
}