通过堆的方式实现
- 时间复杂度:O (n * log k)
空间复杂度:O (n)
function topKFrequent(nums, k) {
const map = new Map();
nums.forEach((n) => {
map.set(n, map.has(n) ? map.get(n) + 1 : 1);
});
const h = new MinHeap();
map.forEach((value, key) => {
h.insert({ value, key });
if (h.size() > k) {
h.pop();
}
});
return h.heap.map((x) => x.key);
}
代码中有两个 forEach 循环,他们的时间复杂度分别是 O (n),而我们知道堆的 insert 和 pop 的时间复杂度分别是 O (log k ) ,所以该函数的时间复杂度为 O (n * log k) k 代表数组中不相同元素的个数。空间复杂度是 O (n),因为主要看 map 中存放不同元素的个数,最坏情况是 nums 数组的长度。
改造了一下堆构造函数
class MinHeap {
constructor() {
this.heap = [];
}
top() {
return this.heap[0];
}
size() {
return this.heap.length;
}
getChildLeftIndex(i) {
return i * 2 + 1;
}
getChildRightIndex(i) {
return i * 2 + 2;
}
getParentIndex(i) {
return (i - 1) >> 1;
}
swap(index1, index2) {
const temp = this.heap[index1];
this.heap[index1] = this.heap[index2];
this.heap[index2] = temp;
}
shiftUp(index) {
if (index === 0) return;
const parentIndex = this.getParentIndex(index);
if (this.heap[parentIndex] && this.heap[parentIndex].value > this.heap[index].value) {
this.swap(parentIndex, index);
this.shiftUp(parentIndex);
}
}
shiftDown(index) {
const leftChildIndex = this.getChildLeftIndex(index);
const rightChildIndex = this.getChildRightIndex(index);
if (this.heap[leftChildIndex] && this.heap[leftChildIndex].value < this.heap[index].value) {
this.swap(leftChildIndex, index);
this.shiftDown(leftChildIndex);
}
if (this.heap[rightChildIndex] && this.heap[rightChildIndex].value < this.heap[index].value) {
this.swap(rightChildIndex, index);
this.shiftDown(rightChildIndex);
}
}
insert(value) {
this.heap.push(value);
this.shiftUp(this.heap.length - 1);
}
pop() {
this.heap[0] = this.heap.pop();
this.shiftDown(0);
}
}