LeetCode 414. Third Maximum Number

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本文介绍了一种解决LeetCode上第三最大数问题的方法，通过一遍扫描的方式找到数组中的第三大数值，如果不存在则返回最大值。该算法的时间复杂度为O(n)，并附带示例说明。

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原题网址：https://leetcode.com/problems/third-maximum-number/

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.

方法：一遍扫描。

public class Solution {
    public int thirdMax(int[] nums) {
        int[] max = new int[3];
        int len = 0;
        for(int num : nums) {
            if (len == 0) {
                max[len++] = num;
            } else if (len == 1) {
                if (max[0] == num) continue;
                if (max[0] < num) {
                    max[1] = max[0];
                    max[0] = num;
                    len++;
                } else {
                    max[len++] = num;
                }
            } else if (len == 2) {
                if (max[0] == num || max[1] == num) continue;
                if (num > max[0]) {
                    max[2] = max[1];
                    max[1] = max[0];
                    max[0] = num;
                    len++;
                } else if (num > max[1]) {
                    max[2] = max[1];
                    max[1] = num;
                    len++;
                } else {
                    max[len++] = num;
                }
            } else {
                if (max[0] == num || max[1] == num || max[2] == num) continue;
                if (num > max[0]) {
                    max[2] = max[1];
                    max[1] = max[0];
                    max[0] = num;
                } else if (num > max[1]) {
                    max[2] = max[1];
                    max[1] = num;
                } else if (num > max[2]) {
                    max[2] = num;
                }
            }
        }
        if (len < 3) return max[0];
        return max[2];
    }
}