動機

寫解題心得才意識到，這就是與two pointer很像

Problem

The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value and the median is the mean of the two middle values.

  • For example, for arr = [2,3,4], the median is 3.
  • For example, for arr = [2,3], the median is (2 + 3) / 2 = 2.5.

Implement the MedianFinder class:

  • MedianFinder() initializes the MedianFinder object.
  • void addNum(int num) adds the integer num from the data stream to the data structure.
  • double findMedian() returns the median of all elements so far. Answers within 10-5 of the actual answer will be accepted.

 

Example 1:

Input[MedianFinder, addNum, addNum, findMedian, addNum, findMedian][[], [1], [2], [], [3], []]Output[null, null, null, 1.5, null, 2.0]ExplanationMedianFinder medianFinder = new MedianFinder();medianFinder.addNum(1);    // arr = [1]medianFinder.addNum(2);    // arr = [1, 2]medianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2)medianFinder.addNum(3);    // arr[1, 2, 3]medianFinder.findMedian(); // return 2.0

 

Constraints:

  • -105 <= num <= 105
  • There will be at least one element in the data structure before calling findMedian.
  • At most 5 * 104 calls will be made to addNum and findMedian.

 

Follow up:

  • If all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?
  • If 99% of all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?

Sol

MaxHeap放小的數字，pop會拿到大的最小 MinHeap放大的數字，pop會拿到小的最大

這樣就能調整大小

class MaxHeap:
    def __init__(self):
        self.hq = []
    def append(self,val):
        heapq.heappush(self.hq,-val)
    def top(self):
        return -self.hq[0]
    def pop(self):
        return -heapq.heappop(self.hq)
    def __len__(self):
        return len(self.hq)
class MinHeap:
    def __init__(self):
        self.hq = []
    def append(self,val):
        heapq.heappush(self.hq,val)
    def top(self):
        return self.hq[0]
    def pop(self):
        return heapq.heappop(self.hq)
    def __len__(self):
        return len(self.hq)
class MedianFinder:

    def __init__(self):
        self.minq, self.maxq = MinHeap(), MaxHeap()
        # premise: 0 <= len(maxq) - len(minq) <= 1
        
    def addNum(self, num: int) -> None:
        if len(self.minq) > 0 and self.minq.top() < num:
            self.minq.append(num)
        else:
            self.maxq.append(num)
        while len(self.maxq) - len(self.minq) > 1 or len(self.maxq) < len(self.minq):
            while len(self.minq) > len(self.maxq):
                self.maxq.append(self.minq.pop())
            while len(self.maxq) - len(self.minq) > 1:
                self.minq.append(self.maxq.pop())
        #print("add",self.minq.hq, self.maxq.hq)

    def findMedian(self) -> float:
        #print("mid",self.minq.hq, self.maxq.hq)
        if len(self.minq) == len(self.maxq):
            return (self.minq.top()+self.maxq.top())/2
        else:
            return self.maxq.top()