## [LeetCode] 1046. Last Stone Weight - 문제풀이

category 알고리즘/LeetCode 2022. 4. 7. 13:34
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# Description

주어진 돌들의 배열에서 아래의 룰에따라 돌을 부셔나갈때 마지막에 남는 돌을 반환하는 문제입니다.

You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:

• If x == y, both stones are destroyed, and
• If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left.

Return the smallest possible weight of the left stone. If there are no stones left, return 0.

Example 1:

``````Input: stones = [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  then that's the value of the last stone.
``````

Example 2:

``````Input: stones = 
Output: 1
``````

Constraints:

• 1 <= stones.length <= 30
• 1 <= stones[i] <= 1000

# Solution 1. PriorityQueue

``````public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> queue = new PriorityQueue<>(Collections.reverseOrder());
for(int i : stones) {
queue.offer(i);
}

while(queue.size() > 1) {
int x = queue.poll();
int y = queue.poll();
if(x > y) {
queue.offer(x-y);
}
}
return queue.isEmpty() ? 0 : queue.poll();
}
``````

우선순위큐(PriorityQueue)를 사용하여 앞에 있는 가장 큰 돌 두개를 꺼낸뒤 둘의 차이를 다시 넣어주고 정렬해 주는 것을 마지막 요소 하나가 남을 때까지 반복해 줍니다.

# Reference

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