Last Stone Weight 最后一块石头的重量
We have a collection of stone , 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.)
输入：[2,7,4,1,8,1]
输出：1
解释：
先选出 7 和 8，得到 1，所以数组转换为 [2,4,1,1,1]，
再选出 2 和 4，得到 2，所以数组转换为 [2,1,1,1]，
接着是 2 和 1，得到 1，所以数组转换为 [1,1,1]，
最后选出 1 和 1，得到 0，最终数组转换为 [1]，这就是最后剩下那块石头的重量。
思路
可以通过大顶堆，将所有数字入堆，每次poll，都是最大的。
做差后大于0 ，就再入堆。
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>((o1,o2)->o2-o1);
for (int stone : stones) {
priorityQueue.offer(stone);
}
while (priorityQueue.size() >1) {
int x = priorityQueue.poll();
int y = priorityQueue.poll();
if (x > y) {
priorityQueue.offer(x - y);
}
}
return priorityQueue.isEmpty() ? 0 : priorityQueue.poll();
}
Tag
heap