合规国际互联网加速 OSASE为企业客户提供高速稳定SD-WAN国际加速解决方案。 广告
Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks. Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle. However, there is a non-negative cooling interval**n**that means between two**same tasks**, there must be at least n intervals that CPU are doing different tasks or just be idle. You need to return the**least**number of intervals the CPU will take to finish all the given tasks. **Example:** ~~~ Input: tasks = ["A","A","A","B","B","B"], n = 2 Output: 8 Explanation: A -> B -> idle -> A -> B -> idle -> A -> B. ~~~ **Note:** 1. The number of tasks is in the range [1, 10000]. 2. The integer n is in the range [0, 100]. ![](https://img.kancloud.cn/26/17/26170998f48aef1f927a09620cc07024_1087x486.png) ``` /** * @param {character[]} tasks * @param {number} n * @return {number} */ var leastInterval = function(tasks, n) { let q = ''; //队列的执行结果 let Q = {}; //对归类进行存储 tasks.forEach(item =>{ if(Q[item]){ Q[item]++; }else{ Q[item] = 1; } }) while(1){ let keys = Object.keys(Q); if(!keys[0]){ break; } let tmp = []; for(let i=0;i<=n;i++){ let max = 0; let key; let pos; keys.forEach((item,idx) =>{ if(Q[item] > max){ max = Q[item]; key = item; pos = idx; } }) 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,'-') } //最后不要出现冷却时间 q = q.replace(/-+$/g,'') return q.length; } ```