LeetCode 1049 Last Stone Weight II - Medium

1049. Last Stone Weight II — Medium

Open on LeetCode

Problem

1049. Last Stone Weight II -- Medium

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose any two rocks 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 smallest possible weight of this stone (the weight is 0 if there are no stones left.)

Example 1:

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

Note:

1 <= stones.length <= 30 1 <= stones[i] <= 100

Solution

# 转换成0-1 Knapsack:
class Solution:
    def lastStoneWeightII(self, stones: List[int]) -> int:
        total = sum(stones)
        Max_weight = int(total/2)
        current = (Max_weight+1)*[0]
        for v in stones:
            for w in range(Max_weight, -1, -1):
                if w-v>=0:
                    current[w] = max(v + current[w-v], current[w])
        return total-2*current[-1]