lc1046. Last Stone Weight

1046. Last Stone Weight Easy

83

8

Favorite

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

Each turn, we choose the two heaviest 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 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 思路：从大到小排序，然后比较头两个元素，如果一样大，就都删除，如果一个比另一个大，插入他们的差的绝对值，while循环，直到只有一个元素或没有

代码：python3

class Solution:
    def lastStoneWeight(self, stones) -> int:
        while len(stones) > 1:
            stones.sort(reverse=True)
            print(stones)
            if stones[0]==stones[1]:
                stones=stones[2:]
            else:
                num=abs(stones[0]-stones[1])
                stones=stones[2:]
                print(stones)
                stones.insert(0,num)
                # print(stones)
        if len(stones)==1:
            return stones[0]
        else:
            return 0
复制代码

转载于:https://juejin.im/post/5d07771a51882559ef78eb27