1046. Last Stone Weight
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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 weightx
is totally destroyed, and the stone of weighty
has new weighty-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
class Solution { public int lastStoneWeight(int[] A) { PriorityQueue<Integer> pq = new PriorityQueue<>((a, b)-> b - a); for (int a : A) pq.offer(a); while (pq.size() > 1) pq.offer(pq.poll() - pq.poll()); return pq.poll(); } }
https://leetcode.com/problems/last-stone-weight/discuss/294956/JavaC%2B%2BPython-Priority-Queue
大神总是能一次次的使我下巴脱臼
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