Saturday, October 15, 2016

LeetCode 414 - Third Maximum Number


https://leetcode.com/problems/third-maximum-number/
Given an array of integers, return the 3rd Maximum Number in this array, if it doesn't exist, return the Maximum Number. The time complexity must be O(n) or less.
http://www.jianshu.com/p/e9dfd2494f41
 public int thirdMax (int[] nums) {
     long first = Long.MIN_VALUE;
     long second = Long.MIN_VALUE;
     long third = Long.MIN_VALUE;
     for (int i:nums){
         if (i>first){
             third = second;
             second = first;
             first = i;
         }else if (i == first)     //必要
             continue;
         else if (i > second){
             third = second;
             second = i;
         }else if (i == second)     //必要
             continue;
         else if (i > third){
             third = i;
         }
     }
     return third == Long.MIN_VALUE ? (int)first : (int)third;
       //method 要求返回类型为int,强制转化一下,narrowing conversion
 }

http://bookshadow.com/weblog/2016/10/09/leetcode-third-maximum-number/
利用变量a, b, c分别记录数组第1,2,3大的数字
遍历一次数组即可,时间复杂度O(n)

def thirdMax(self, nums): """ :type nums: List[int] :rtype: int """ a = b = c = None for n in nums: if n > a: a, b, c = n, a, b elif a > n > b: b, c = n, b elif b > n > c: c = n return c if c is not None else a

https://discuss.leetcode.com/topic/65119/java-solution-using-treeset
    public final int N = 3;
    public int thirdMax(int[] nums) {
        if (nums.length == 0) return 0;

        TreeSet<Integer> set = new TreeSet<Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (set.contains(nums[i])) continue;
            if (set.size() < N || nums[i] > set.first()) {
                if (set.size() == N) set.remove(set.first());
                set.add(nums[i]);
            }
        }
        return set.size() == N ? set.first() : set.last();
    }

http://www.cnblogs.com/grandyang/p/5983113.html
下面这种方法利用了set的自动排序和自动去重复项的特性,很好的解决了问题,对于遍历到的数字,加入set中,重复项就自动去掉了,如果此时set大小大于3个了,那么我们把set的第一个元素去掉,也就是将第四大的数字去掉,那么就可以看出set始终维护的是最大的三个不同的数字,最后遍历结束后,我们看set的大小是否为3,是的话就返回首元素,不是的话就返回尾元素

    int thirdMax(vector<int>& nums) {
        set<int> s;
        for (int num : nums) {
            s.insert(num);
            if (s.size() > 3) {
                s.erase(s.begin());
            }
        }
        return s.size() == 3 ? *s.begin() : *s.rbegin();
    }
https://discuss.leetcode.com/topic/63086/java-priorityqueue-o-n-o-1
    public int thirdMax(int[] nums) {
       PriorityQueue<Integer> pq = new PriorityQueue<>();
       Set<Integer> set = new HashSet<>();
       for(int n : nums) {
           if(set.add(n)) {
               pq.offer(n);
               if(pq.size() > 3 ) pq.remove(pq.poll());
           }
       }
       if(pq.size() == 2) pq.poll();
       return pq.peek();
    }


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