[LeetCode]414 Third Maximum Number

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本文介绍了一个高效的算法，用于从整数数组中找出第三大的数。若不存在第三大的数，则返回最大的数。通过示例展示了算法如何工作，并提供了一个Java实现方案。

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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.

public class Solution {
    public int thirdMax(int[] nums) {
        int first = -1;
        int second = -1;
        int third = -1;
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] == first || nums[i] == second || nums[i] == third)
                continue;
            if(first == -1 || nums[i] > first) {
                third = second;
                second = first;
                first = nums[i];
            } else if(second == -1 || nums[i] > second) {
                third = second;
                second = nums[i];
            } else if(third == -1 || nums[i] > third) {
                third = nums[i];
            }
        }
        return third == -1 ? first : third;
    }
}

如果数组中有-1出现，刚会出现错误，应该用Integer，默认值为null为表示

Java neat and easy understand solution, O(n) time, O(1) space

    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;
    }