LeetCode 1046. Last Stone Weight

Posted 2020-11-21 Dylan_Java_NYC

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 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. 1 <= stones.length <= 30
2. 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 }

跟上Last Stone Weight II.