原题链接在这里:https://leetcode.com/problems/last-stone-weight/
题目:
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
题解:
Put all stones into max heap.
While heap size >= 2, poll top 2 elements and get diff. If diff is larger than 0, add it back to heap.
Time Complexity: O(nlogn). n = stones.length. while loop could run for maximumn n-1 times.
Space: O(n).
AC Java:
1 class Solution { 2 public int lastStoneWeight(int[] stones) { 3 if(stones == null || stones.length == 0){ 4 return 0; 5 } 6 7 PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder()); 8 for(int stone : stones){ 9 maxHeap.add(stone); 10 } 11 12 while(maxHeap.size() > 1){ 13 int x = maxHeap.poll(); 14 int y = maxHeap.poll(); 15 int diff = x-y; 16 if(diff > 0){ 17 maxHeap.add(diff); 18 } 19 } 20 21 return maxHeap.isEmpty() ? 0 : maxHeap.peek(); 22 } 23 }
原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11785085.html
时间: 2024-11-05 19:40:40