题目链接：https://leetcode-cn.com/problems/task-scheduler/
思路：数量多的先安排任务，少的后安排，每次仅仅采用每类任务个数减1的方式
class Solution { public: struct node { node():ch('.'),count(0){} node(char c,int n):ch(c),count(n) { } bool friend operator<(node x,node y) { if(x.count!=y.count) return x.count<y.count; else return x.ch<y.ch; } char ch; int count; }; int leastInterval(vector<char>& tasks, int n) { unordered_map<char,int>mapp; priority_queue<node>pri_que; int result=0; for(auto ch:tasks) { if(mapp.count(ch)) { mapp[ch]++; } else { mapp[ch]=1; } } for(auto pair_temp:mapp) { pri_que.push(node(pair_temp.first,pair_temp.second)); } while(!pri_que.empty()) { vector<node>vec_temp; for(int i=0;i<n+1;i++) { if(!pri_que.empty()) { node lef_node=pri_que.top(); pri_que.pop(); lef_node.count-=1; cout<<lef_node.ch<<" "<<lef_node.count<<endl; if(lef_node.count!=0) { vec_temp.push_back(lef_node); } } result++; if(pri_que.empty()&&vec_temp.empty())break; } for(auto temp_node:vec_temp) { pri_que.push(temp_node); } } return result; } };
