leetcode题解-414. Third Maximum Number

题目：

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: 1Explanation: The third maximum is 1.Example 2:Input: [1, 2]Output: 2Explanation: The third maximum does not exist, so the maximum (2) is returned instead.Example 3:Input: [2, 2, 3, 1]Output: 1Explanation: Note that the third maximum here means the third maximum distinct number.Both numbers with value 2 are both considered as second maximum.

本题是要寻找数组中第三大的数，我们的第一种思路就是使用三个数来保存最大的三个数，便利数组的同时进行修改即可。代码入下：

    public int thirdMax1(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;    }

此外，我们还可以使用一个变量来保存数字，一个变量来表示当前保存的是第几大的数字。首先将数组进行排序，然后从数组的最后开始遍历，即从最大的数字开始遍历。代码入下：

    public int thirdMax(int[] nums) {        int res=Integer.MIN_VALUE, count=0;        Arrays.sort(nums);        for(int i=nums.length-1; i>=0; i--){            if(res != nums[i]){                res = nums[i];                count++;                if(count == 3)                    break;            }        }        if(count<3)            return nums[nums.length-1];        return res;    }

这两种方法思路差不多，效率也很相近。此外我还看到一种使用priorityQueue的方法，使用优先级队列来保存最大的三个数字。但是因为其本身效率很低，所以代码耗时较长。代码入下：

    public int thirdMax2(int [] nums){        PriorityQueue<Integer> pq = new PriorityQueue<>();        Set<Integer> set = new HashSet<>();        for(int num : nums){            if(!set.contains(num)){                pq.offer(num);                set.add(num);                if(pq.size() > 3)                    set.remove(pq.poll());            }        }        if(pq.size()<3)            while(pq.size()>1)                pq.poll();        return pq.peek();    }