隊(duì)列 (常用數(shù)據(jù)結(jié)構(gòu)之一)
隊(duì)列是一種特殊的線(xiàn)性表吴菠,特殊之處在于它只允許在表的前端(front)進(jìn)行刪除操作者填,而在表的后端(rear)進(jìn)行插入操作,和棧一樣做葵,隊(duì)列是一種操作受限制的線(xiàn)性表占哟。進(jìn)行插入操作的端稱(chēng)為隊(duì)尾,進(jìn)行刪除操作的端稱(chēng)為隊(duì)頭。隊(duì)列中沒(méi)有元素時(shí)榨乎,稱(chēng)為空隊(duì)列怎燥。
隊(duì)列的數(shù)據(jù)元素又稱(chēng)為隊(duì)列元素。在隊(duì)列中插入一個(gè)隊(duì)列元素稱(chēng)為入隊(duì)蜜暑,從隊(duì)列中刪除一個(gè)隊(duì)列元素稱(chēng)為出隊(duì)铐姚。因?yàn)殛?duì)列只允許在一端插入,在另一端刪除肛捍,所以只有最早進(jìn)入隊(duì)列的元素才能最先從隊(duì)列中刪除谦屑,故隊(duì)列又稱(chēng)為先進(jìn)先出(FIFO—first in first out)線(xiàn)性表
題目:622. 設(shè)計(jì)循環(huán)隊(duì)列
思路:用數(shù)組實(shí)現(xiàn)隊(duì)列結(jié)構(gòu),當(dāng)隊(duì)列全滿(mǎn)的時(shí)候篇梭,把尾指針指向隊(duì)首
export default class MyCircularQueue {
constructor(k) {
// 用來(lái)保存數(shù)據(jù)長(zhǎng)度為k的數(shù)據(jù)結(jié)構(gòu)
this.list = new Array(k);
// 隊(duì)首指針
this.front = 0;
// 隊(duì)尾指針
this.rear = 0;
// 隊(duì)列長(zhǎng)度
this.max = k;
}
enQueue(num) {
if (this.isFull()) {
return false;
} else {
this.list[this.rear] = num;
this.rear = (this.rear + 1) % this.max;
return true;
}
}
deQueue() {
let v = this.list[this.front];
this.list[this.front] = '';
this.front = (this.front + 1) % this.max;
return v;
}
isEmpty() {
return this.front === this.rear && !this.list[this.front];
}
isFull() {
return this.front === this.rear && !!this.list[this.front];
}
Front() {
return this.list[this.front];
}
Rear() {
let rear = this.rear - 1;
return this.list[rear < 0 ? this.max - 1 : rear];
}
}
題目: 621. 任務(wù)調(diào)度器
- 先按次數(shù)分類(lèi)任務(wù)
- 然后按次數(shù)從大到小找到n+1個(gè)任務(wù)氢橙,循環(huán)做這步,直到任務(wù)找完
function leastInterval(tasks, n) {
let q = '';
let Q = tasks.reduce((Q, item) => {
if (Q[item]) {
Q[item]++;
} else {
Q[item] = 1;
}
return Q;
}, {});
while (1) {
// 獲取不同的任務(wù)列表
let keys = Object.keys(Q);
// 如果已經(jīng)沒(méi)有任務(wù)恬偷,就退出循環(huán)處理
if (!keys[0]) {
break;
}
// 處理一組數(shù)據(jù)
let tmp = [];
for (let i = 0; i <= n; i++) {
let max = 0;
let key;
let pos;
// 從所有的任務(wù)中找到未處理數(shù)最大的優(yōu)先安排
keys.forEach((item, idx) => {
if (Q[item] > max) {
max = Q[item];
key = item;
pos = idx;
}
})
// 找到可以就放進(jìn)tmp數(shù)組中悍手,否則跳出while循環(huán)
if (key) {
tmp.push(key);
keys.splice(pos, 1);
Q[key]--;
if (Q[key] < 1) {
delete Q[key];
}
} else {
break;
}
}
q += tmp.join('').padEnd(n + 1, '-');
}
q = q.replace(/-+$/g, '');
return q.length;
}