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1046.最后一块石头的重量

链接:1046.最后一块石头的重量
难度:Easy
标签:数组、堆(优先队列)
简介:有一堆石头,每块石头的重量都是正整数。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。

题解 1 - typescript

  • 编辑时间:2020-12-30
  • 执行用时:104ms
  • 内存消耗:40MB
  • 编程语言:typescript
  • 解法介绍:构建堆。
var lastStoneWeight = function (stones) {
const pq = new MaxPriorityQueue();
for (const stone of stones) {
pq.enqueue('x', stone);
}
while (pq.size() > 1) {
const a = pq.dequeue()['priority'];
const b = pq.dequeue()['priority'];
if (a > b) {
pq.enqueue('x', a - b);
}
}
return pq.isEmpty() ? 0 : pq.dequeue()['priority'];
};

题解 2 - typescript

  • 编辑时间:2021-04-11
  • 执行用时:92ms
  • 内存消耗:39.1MB
  • 编程语言:typescript
  • 解法介绍:构建堆。
class Heap<T = number> {
private arr: T[] = [];
get isEmpty() {
return this.size === 0;
}
get size() {
return this.arr.length;
}
get top() {
return this.arr[0];
}
constructor(private compare: (t1: T, t2: T) => number) {}
add(num: T): void {
this.arr.push(num);
this.shiftUp(this.size - 1);
}
remove(): T {
const num = this.arr.shift()!;
if (this.size) {
this.arr.unshift(this.arr.pop()!);
this.shiftDown(0);
}
return num;
}
private shiftUp(index: number): void {
if (index === 0) return;
const parentIndex = (index - 1) >> 1;
if (this.compare(this.arr[index], this.arr[parentIndex]) > 0) {
[this.arr[index], this.arr[parentIndex]] = [this.arr[parentIndex], this.arr[index]];
this.shiftUp(parentIndex);
}
}
private shiftDown(index: number): void {
let childrenIndex = index * 2 + 1;
if (childrenIndex > this.size - 1) return;
if (
childrenIndex + 1 <= this.size - 1 &&
this.compare(this.arr[childrenIndex + 1], this.arr[childrenIndex]) > 0
) {
childrenIndex++;
}
if (this.compare(this.arr[childrenIndex], this.arr[index]) > 0) {
[this.arr[childrenIndex], this.arr[index]] = [this.arr[index], this.arr[childrenIndex]];
this.shiftDown(childrenIndex);
}
}
*[Symbol.iterator](): IterableIterator<T> {
for (const t of this.arr) {
yield t;
}
}
}
function lastStoneWeight(stones: number[]): number {
const heap = new Heap((t1, t2) => t1 - t2);
stones.forEach(v => heap.add(v));
while (heap.size > 1) {
const s1 = heap.remove();
const s2 = heap.remove();
if (s1 === s2) continue;
heap.add(Math.abs(s1 - s2));
}
return heap.size === 0 ? 0 : heap.top;
}