Open azl397985856 opened 1 year ago
class Solution {
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a, b) -> b - a);
for (int stone : stones) {
pq.offer(stone);
}
while (pq.size() > 1) {
int a = pq.poll();
int b = pq.poll();
if (a > b) {
pq.offer(a - b);
}
}
return pq.isEmpty() ? 0 : pq.poll();
}
}
class Solution {
public:
int lastStoneWeight(vector<int>& stones) {
sort(stones.begin(),stones.end());
int n=stones.size();
priority_queue<int,vector<int>,less<int>> pq;
for(int s:stones){
pq.push(s);
}
while(pq.size()!=1){
// cout<<pq.size()<<endl;
int a=pq.top();
pq.pop();
int b=pq.top();
pq.pop();
pq.push(a-b);
// cout<<a<<" "<<b<<endl;
if(pq.size()==1)break;
}
return pq.top();
}
};
class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
# choose 最重的两个
stones = [-x for x in stones]
heapq.heapify(stones)
while len(stones) >= 2:
x = heappop(stones)
y = heappop(stones)
if x == y:
continue
else:
heappush(stones, -abs(x - y))
if len(stones) >= 1:
return -stones[0]
return 0
/**
TC: O(NlogN) SC: O(N)
*/
class Solution1 {
public int lastStoneWeight(int[] stones) {
// MaxHeap
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> Integer.compare(b, a));
for (int x : stones)
pq.offer(x);
while (pq.size() > 1) {
int a = pq.poll();
int b = pq.poll();
if (a > b)
pq.offer(a - b);
}
return pq.isEmpty() ? 0: pq.poll();
}
}
大顶堆
/**
* @param {number[]} stones
* @return {number}
*/
var lastStoneWeight = function(stones) {
const maxQ = new MaxPriorityQueue();
for (let i = 0; i < stones.length; ++i) {
maxQ.enqueue(stones[i])
}
while(maxQ.size()>1) {
const one = maxQ.dequeue().element;
const two = maxQ.dequeue().element;
if (one !== two) {
maxQ.enqueue(one-two);
}
}
return maxQ.dequeue()?.element ?? 0;
};
时间O(nlogn) 空间O(n)
class Solution {
public:
int lastStoneWeight(vector<int>& stones) {
int n=stones.size();
if(n==1) return stones[0];
for(int i=n-1;i>0;i--){
sort(stones.begin(),stones.end());
if(stones[i-1]==0) continue;
if(stones[i]==stones[i-1]) {
stones[i]=stones[i-1]=0;
}
else{
stones[i-1]=stones[i]-stones[i-1];
stones[i]=0;
}
i=n;
}
return stones[n-1];
}
};
class Solution {
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> queue = new PriorityQueue<>((a, b) -> b - a);
for(int i = 0; i < stones.length; i++){
queue.add(stones[i]);
}
while(queue.size() > 1){
int y = queue.poll();
int x = queue.poll();
if(x == y){
continue;
}else{
queue.add(y - x);
}
}
return queue.size() == 1 ? queue.poll() : 0;
}
}
code
public int lastStoneWeight(int[] stones) {
var pq = new PriorityQueue<Integer>(Comparator.reverseOrder());
for (int stone : stones) pq.add(stone);
while (pq.size() > 1) {
int y = pq.poll();
int x = pq.poll();
if (y > x) pq.offer(y - x);
}
return pq.isEmpty() ? 0 : pq.poll();
}
class Solution: def lastStoneWeight(self, stones: List[int]) -> int: h = [-x for x in stones] heapq.heapify(h) while len(h) > 1 and h[0] != 0: heapq.heappush(h, heapq.heappop(h) - heapq.heappop(h)) return -h[0]
class Solution {
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> heap = new PriorityQueue<>(stones.length, new Comparator<Integer>(){
@Override
public int compare(Integer o1, Integer o2) {
return o2 - o1;
}
});
for(int num : stones) {
heap.offer(num);
}
while (heap.size() > 1) {
int largest = heap.poll();
int large = heap.poll();
if (largest > large) {
heap.offer(largest - large);
}
}
return heap.isEmpty() ? 0 : heap.poll();
}
}
class Solution: def lastStoneWeight(self, stones: List[int]) -> int: while len(stones)>=2: stones.sort() stones.append(stones.pop()-stones.pop()) return stones[0]
class Solution {
public:
int lastStoneWeight(vector<int>& stones) {
priority_queue<int> q;
for (int s: stones) {
q.push(s);
}
while (q.size() > 1) {
int a = q.top();
q.pop();
int b = q.top();
q.pop();
if (a > b) {
q.push(a - b);
}
}
return q.empty() ? 0 : q.top();
}
};
1046.最后一块石头的重量
入选理由
暂无
题目地址
https://leetcode-cn.com/problems/last-stone-weight/
前置知识
题目描述
每一回合,从中选出两块 最重的 石头,然后将它们一起粉碎。假设石头的重量分别为 x 和 y,且 x <= y。那么粉碎的可能结果如下:
如果 x == y,那么两块石头都会被完全粉碎; 如果 x != y,那么重量为 x 的石头将会完全粉碎,而重量为 y 的石头新重量为 y-x。 最后,最多只会剩下一块石头。返回此石头的重量。如果没有石头剩下,就返回 0。
示例:
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。
提示:
1 <= stones.length <= 30 1 <= stones[i] <= 1000