zoukankan      html  css  js  c++  java
  • [LeetCode 1167] Minimum Cost to Connect Sticks

    You have some sticks with positive integer lengths.

    You can connect any two sticks of lengths X and Y into one stick by paying a cost of X + Y.  You perform this action until there is one stick remaining.

    Return the minimum cost of connecting all the given sticks into one stick in this way.

     

    Example 1:

    Input: sticks = [2,4,3]
    Output: 14
    

    Example 2:

    Input: sticks = [1,8,3,5]
    Output: 30
    

     

    Constraints:

    • 1 <= sticks.length <= 10^4
    • 1 <= sticks[i] <= 10^4

    This is a direct application of the Huffman Coding algorithm. In this problem, each stick's length is the equivalence of character frequency, the goal here is to merge all sticks into one stick with the minimum cost, this is essentially the same with building a Huffman tree with minimum cost. Each merge operation merges two meta-sticks(two subtrees) into one subtree with the introduction of a new internal tree node. The gist of the Huffman Coding is to greedily pick the 2 least frequent characters and merge them, then recursively solve a smaller problem with 1 fewer character. In this problem, this translates into picking the 2 least stick lengths.

    There are two efficient ways of implementing the Huffman Coding algorithm.

    Solution 1. PriorityQueue

    Since we conduct repeated finding minimum operations, PriorityQueue suits well for such application. 

    class Solution {
        public int connectSticks(int[] sticks) {
            PriorityQueue<Integer> minPq = new PriorityQueue<>();
            for(int v : sticks) {
                minPq.add(v);
            }
            int cost = 0;
            while(minPq.size() > 1) {
                int v1 = minPq.poll();
                int v2 = minPq.poll();
                cost += (v1 + v2);
                minPq.add(v1 + v2);
            }
            return cost;
        }
    }

    Solution 2. Sorting + Two Queue

    Alternatively, we can preprocess the input by sorting it, then manage two queues. The first queue initially stores all input. The second queue is used to store all the meta sticks' length after merging operations. As long as there is more than 1 element left in these 2 queues combined, we repeatedly pick the 2 smallest stick lengths and merge them.

    class Solution {
        public int connectSticks(int[] sticks) {
            Arrays.sort(sticks);
            ArrayDeque<Integer> q1 = new ArrayDeque<>();
            ArrayDeque<Integer> q2 = new ArrayDeque<>();
            for(int v : sticks) {
                q1.addLast(v);
            }
            
            int cost = 0;
            while(!(q1.size() == 1 && q2.size() == 0 || q1.size() == 0 && q2.size() == 1)) {
                int c1 = Integer.MAX_VALUE, c2 = Integer.MAX_VALUE;
                if(q1.size() > 0 && (q2.size() == 0 || q1.peekFirst() <= q2.peekFirst())) {
                    c1 = q1.pollFirst();
                }
                else {
                    c1 = q2.pollFirst();
                }
                if(q1.size() > 0 && (q2.size() == 0 || q1.peekFirst() <= q2.peekFirst())) {
                    c2 = q1.pollFirst();
                }
                else {
                    c2 = q2.pollFirst();
                }
                cost += (c1 + c2);
                q2.addLast(c1 + c2);
            }
            return cost;
        }
    }
  • 相关阅读:
    oracle中的exists 和not exists 用法详解
    再次谈谈easyui datagrid 的数据加载
    oracle之trunc(sysdate)
    小菜学习设计模式(五)—控制反转(Ioc)
    vim实用技巧
    003_Linux的Cgroup<实例详解>
    systemd在各个linux发行版的普及
    (部署新java程序,程序报错,需copy的一个包)——java使用siger 获取服务器硬件信息
    中国科学院国家授时中心
    Linux时间同步配置方法
  • 原文地址:https://www.cnblogs.com/lz87/p/12231837.html
Copyright © 2011-2022 走看看