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.
Solution:
Java
class Solution { public int thirdMax(int[] nums) { long[] max = {Long.MIN_VALUE, Long.MIN_VALUE, Long.MIN_VALUE}; int count = 0; for (int num : nums) { for (int j = 0; j < 3; j++) { if (max[j] > num) continue; else if (max[j] == num) break; int k = j; long temp1, temp2; temp1 = num; count++; while (k < 3) { temp2 = max[k]; max[k] = temp1; temp1 = temp2; k++; } break; } } System.out.println(Integer.MIN_VALUE); return (count >= 3) ? (int) max[2] : (int) max[0]; } }