動機

練heapq

Problem

We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is:

  • If x == y, both stones are totally destroyed;
  • If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.

At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)

 

Example 1:

Input: [2,7,4,1,8,1]Output: 1Explanation: We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.

 

Note:

  1. 1 <= stones.length <= 30
  2. 1 <= stones[i] <= 1000

Sol

拿前二大的相減，再塞回去

class Solution:
    def lastStoneWeight(self, q: List[int]) -> int:
        q = [-x for x in q]
        heapify(q)
        while len(q) > 1:
            a = -heappop(q)
            b = -heappop(q)
            if a != b:
                heappush(q, -a+b)
        return -q[0] if len(q) > 0 else 0