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Formatted question description: https://leetcode.ca/all/1046.html
1046. Last Stone Weight (Easy)
We have a collection of stones, each stone has a positive integer weight.
Each turn, we choose the two heaviest stones 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
Solution 1. Max-root Heap
// OJ: https://leetcode.com/problems/last-stone-weight/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
int lastStoneWeight(vector<int>& A) {
priority_queue<int> pq(begin(A), end(A));
while (pq.size() > 1) {
int a = pq.top();
pq.pop();
int b = pq.top();
pq.pop();
pq.push(a - b);
}
return pq.top();
}
};
-
class Solution { public int lastStoneWeight(int[] stones) { PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(new Comparator<Integer>() { public int compare(Integer num1, Integer num2) { return num1 == null || num2 == null ? 0 : num2 - num1; } }); for (int stone : stones) priorityQueue.offer(stone); while (priorityQueue.size() > 1) { int stone1 = priorityQueue.poll(); int stone2 = priorityQueue.poll(); if (stone1 > stone2) priorityQueue.offer(stone1 - stone2); } return priorityQueue.size() == 0 ? 0 : priorityQueue.poll(); } } ############ class Solution { public int lastStoneWeight(int[] stones) { PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a); for (int v : stones) { q.offer(v); } while (q.size() > 1) { int y = q.poll(); int x = q.poll(); if (x != y) { q.offer(y - x); } } return q.isEmpty() ? 0 : q.poll(); } }
-
// OJ: https://leetcode.com/problems/last-stone-weight/ // Time: O(NlogN) // Space: O(N) class Solution { public: int lastStoneWeight(vector<int>& A) { priority_queue<int> pq(begin(A), end(A)); while (pq.size() > 1) { int a = pq.top(); pq.pop(); int b = pq.top(); pq.pop(); int r = abs(a - b); if (r) pq.push(r); } return pq.size() ? pq.top() : 0; } };
-
class Solution: def lastStoneWeight(self, stones: List[int]) -> int: h = [-v for v in stones] heapify(h) while len(h) > 1: x = heappop(h) y = heappop(h) if x != y: heappush(h, x - y) return 0 if len(h) == 0 else -h[0] ############ class Solution(object): def lastStoneWeight(self, stones): """ :type stones: List[int] :rtype: int """ stones = map(lambda x : -x, stones) heapq.heapify(stones) while len(stones) > 1: x = heapq.heappop(stones) if stones: y = heapq.heappop(stones) if x != y: heapq.heappush(stones, -abs(x - y)) return 0 if not stones else -stones[0]
-
func lastStoneWeight(stones []int) int { q := &hp{stones} heap.Init(q) for q.Len() > 1 { x, y := q.pop(), q.pop() if x != y { q.push(x - y) } } if q.Len() > 0 { return q.IntSlice[0] } return 0 } type hp struct{ sort.IntSlice } func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] } func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) } func (h *hp) Pop() interface{} { a := h.IntSlice v := a[len(a)-1] h.IntSlice = a[:len(a)-1] return v } func (h *hp) push(v int) { heap.Push(h, v) } func (h *hp) pop() int { return heap.Pop(h).(int) }
-
/** * @param {number[]} stones * @return {number} */ var lastStoneWeight = function (stones) { const pq = new MaxPriorityQueue(); for (const v of stones) { pq.enqueue(v); } while (pq.size() > 1) { const x = pq.dequeue()['priority']; const y = pq.dequeue()['priority']; if (x != y) { pq.enqueue(x - y); } } return pq.isEmpty() ? 0 : pq.dequeue()['priority']; };
-
function lastStoneWeight(stones: number[]): number { const pq = new MaxPriorityQueue(); for (const x of stones) { pq.enqueue(x); } while (pq.size() > 1) { const y = pq.dequeue().element; const x = pq.dequeue().element; if (x !== y) { pq.enqueue(y - x); } } return pq.isEmpty() ? 0 : pq.dequeue().element; }