给你一个整数数组 nums 和一个整数 k ,请你返回其中出现频率前 k 高的元素。你可以按 任意顺序 返回答案。
示例 1:
输入:nums = [1,1,1,2,2,3], k = 2
输出:[1,2]
示例 2:
输入:nums = [1], k = 1
输出:[1]
这道题我们用到了大顶堆,我们先用map将数字的个数进行统计,然后加入大顶堆中,再返回k次堆顶元素即可。
var topKFrequent = function (nums, k) {
let map = new Map();
let heap = new Heap((a, b) => a.val < b.val)
let res = []
for (let i = 0; i < nums.length; i++) {
map.set(nums[i], map.has(nums[i]) ? map.get(nums[i]) + 1 : 1)
}
map.forEach((val, key) => {
heap.push({
key, val
})
})
for (let i = 0; i < k; i++) {
res.push(heap.pop().key)
}
return res
};
通用堆写法
class Heap {
constructor(cmp = "large") {
if (cmp == "large") {
this.cmp = this.large;
} else if (cmp == "small") {
this.cmp = this.small
} else {
this.cmp = cmp
}
this.res = [];
this.cnt = 0;
}
push(val) {
this.cnt++;
this.res.push(val)
this.shiftUp(this.cnt - 1)
}
pop() {
this.cnt--;
const res = this.res[0]
const pop = this.res.pop()
if (this.cnt) {
this.res[0] = pop
this.shiftDown(0)
}
return res
}
shiftUp(i) {
if (i === 0) return
const par = this.getParentIndex(i)
if (this.cmp(this.res[par], this.res[i])) {
this.swap(par, i)
this.shiftUp(par)
}
}
shiftDown(i) {
const l = this.getLeftIndex(i)
const r = this.getRightIndex(i)
if (l < this.cnt && this.cmp(this.res[i], this.res[l])) {
this.swap(i, l)
this.shiftDown(l)
}
if (r < this.cnt && this.cmp(this.res[i], this.res[r])) {
this.swap(i, r)
this.shiftDown(r)
}
}
getParentIndex(i) {
return (i - 1) >> 1
}
getLeftIndex(i) {
return i * 2 + 1
}
getRightIndex(i) {
return i * 2 + 2
}
large = (a, b) => a < b
small = (a, b) => a > b;
swap = (i, j) => [this.res[i], this.res[j]] = [this.res[j], this.res[i]];
top = () => this.res[0];
size = () => this.cnt;
isEmpty = () => this.cnt === 0
}