1046. Last Stone Weight-最后一块石头的重量

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: 1
Explanation:
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 <= stones.length <= 30
• 1 <= stones[i] <= 1000

Solution in python:

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        stones.sort()
        while len(stones) > 1:
            m1 = stones.pop()
            m2 = stones.pop()
            if m1 == m2:
                continue
            else:
                i = 0
                new = m1 - m2
                while i < len(stones) and stones[i] < new:
                    i += 1 
                stones.insert(i, new)

        if stones:
            return stones[0]
        else:
            return 0