代码(Python3) class Solution: def lastStoneWeight(self, stones: List[int]) -> int: # 由于 heapq 默认是最小堆,所以这里取相反数进行转换 heap: List[int] = [-num for num in stones] # 通过数组直接建立堆 heapq.heapify(heap) # 至少还有两个数字时,继续循环处理 while len(heap) > 1: # ...
最后判断队列是否为空,为空就返回0,不为空就返回队首的元素值。 publicintlastStoneWeight3(int[] stones){ PriorityQueue<Integer> pq =newPriorityQueue<>(Comparator.reverseOrder());for(intstone : stones) { pq.offer(stone); }while(pq.size() >1) {intnum=pq.poll() - pq.poll();if(num >0)...
来源:力扣(LeetCode)链接:https://leetcode-cn.com/problems/last-stone-weight 提示 为了实现全局最小,首先从实现局部最小开始吧 本题关键字:贪心算法,优先队列 解析 思路 对于每一个stone, s,为了使相减结果最小,我们必然要寻找重量最接近s的另一个stone,且由于选大选小关系不方便判断,可以尝试从最大的s开始...
LeetCode 1049. Last Stone Weight II LeetCode 1049. Last Stone Weight II (最后一块石头的重量 II) 题目 链接 https://leetcode.cn/problems/last-stone-weight-ii/ 问题描述 有一堆石头,用整数数组 stones 表示。其中 stones[i] 表示第 i 块石头的重量。 每一回合,从中选出任意两块石头,然后将它们一...
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 any two stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is: ...
packageleetcodefunclastStoneWeightII(stones[]int)int{sum:=0for_,v:=rangestones{sum+=v}n,C,dp:=len(stones),sum/2,make([]int,sum/2+1)fori:=0;i<=C;i++{ifstones[0]<=i{dp[i]=stones[0]}else{dp[i]=0}}fori:=1;i<n;i++{forj:=C;j>=stones[i];j--{dp[j]=max(dp[...
pq.offer(pq.poll()-pq.poll());returnpq.poll(); } } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. https://leetcode.com/problems/last-stone-weight/discuss/294956/JavaC%2B%2BPython-Priority-Queue 大神总是能一次次的使我下巴脱臼
1049. Last Stone Weight II刷题笔记 class Solution: def lastStoneWeightII(self, stones: List[int]) -> int: sumweight = sum(stones) target = sumweight//2 dp = [0]*(target+1) for i in range(len(stones)): for j in range(target,stones[i]-1,-1):...
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classSolution{public:intlastStoneWeight(vector<int>& stones){ priority_queue<int> q;for(intstone : stones) q.push(stone);while(q.size() >1) {intfirst = q.top(); q.pop();intsecond = q.top(); q.pop();if(first > second) q.push(first - second); ...