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414. Third Maximum Number

414. Third Maximum Number

作者: matrxyz | 来源:发表于2018-01-16 12:39 被阅读0次

    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:遍历

    思路:

    Time Complexity: O(N) Space Complexity: O(1)

    Solution Code:

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

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