Question

1046. Last Stone Weight

中文文档

Description

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 weight of the last remaining 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 [1] then that's the value of the last stone.

Example 2:

Input: stones = [1]
Output: 1

 

Constraints:

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

Solutions

Solution 1

class Solution:
  def lastStoneWeight(self, stones: List[int]) -> int:
    h = [-x for x in stones]
    heapify(h)
    while len(h) > 1:
      y, x = -heappop(h), -heappop(h)
      if x != y:
        heappush(h, x - y)
    return 0 if not h else -h[0]

class Solution {
  public int lastStoneWeight(int[] stones) {
    PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
    for (int x : stones) {
      q.offer(x);
    }
    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();
  }
}

class Solution {
public:
  int lastStoneWeight(vector<int>& stones) {
    priority_queue<int> pq;
    for (int x : stones) {
      pq.push(x);
    }
    while (pq.size() > 1) {
      int y = pq.top();
      pq.pop();
      int x = pq.top();
      pq.pop();
      if (x != y) {
        pq.push(y - x);
      }
    }
    return pq.empty() ? 0 : pq.top();
  }
};

func lastStoneWeight(stones []int) int {
  q := &hp{stones}
  heap.Init(q)
  for q.Len() > 1 {
    y, x := q.pop(), q.pop()
    if x != y {
      q.push(y - x)
    }
  }
  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 any)    { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
  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) }

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;
}

/**
 * @param {number[]} stones
 * @return {number}
 */
var lastStoneWeight = function (stones) {
  const pq = new MaxPriorityQueue();
  for (const x of stones) {
    pq.enqueue(x);
  }
  while (pq.size() > 1) {
    const y = pq.dequeue()['priority'];
    const x = pq.dequeue()['priority'];
    if (x != y) {
      pq.enqueue(y - x);
    }
  }
  return pq.isEmpty() ? 0 : pq.dequeue()['priority'];
};