- Last Stone Weight Easy
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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