README_EN.md

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

Python3

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]

Java

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();
    }
}

C++

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

Go

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) }

JavaScript

/**
 * @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'];
};

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