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数据结构与算法——前缀树和贪心算法(1)

时间:2020-01-23 16:44:51      阅读:106      评论:0      收藏:0      [点我收藏+]

标签:字符串   --   拼接   答案   删除   tran   ring   als   c++   

介绍前缀树

何为前缀树?如何生成前缀树?

例子:一个字符串类型的数组arrl,另一个字符串类型的数组arr2。arr2中有哪些字符,是arr 1中 出现的?请打印。arr2中有哪些字符,是作为arr 1中某个字符串前缀出现的?请打印。arr2 中有哪些字符,是作为arr1中某个字符串前缀出现的?请打印arr2中出现次数最大的前缀。

public class TrieTree {

    public static class TrieNode {
        public int path;
        public int end;
        public TrieNode[] nexts;

        public TrieNode() {
            path = 0;
            end = 0;
            nexts = new TrieNode[26];
        }
    }

    public static class Trie {
        private TrieNode root;

        public Trie() {
            root = new TrieNode();
        }

        public void insert(String word) {
            if (word == null) {
                return;
            }
            char[] chs = word.toCharArray();
            TrieNode node = root;
            int index = 0;
            for (int i = 0; i < chs.length; i++) {
                index = chs[i] - 'a';
                if (node.nexts[index] == null) {
                    node.nexts[index] = new TrieNode();
                }
                node = node.nexts[index];
                node.path++;
            }
            node.end++;
        }

        public void delete(String word) {
            if (search(word) != 0) {  //确定树中确定加入过word,才删除
                char[] chs = word.toCharArray();
                TrieNode node = root;
                int index = 0;
                for (int i = 0; i < chs.length; i++) {
                    index = chs[i] - 'a';
                    if (--node.nexts[index].path == 0) {  //C++要遍历到底去析构
                        node.nexts[index] = null;
                        return;
                    }
                    node = node.nexts[index];
                }
                node.end--;
            }
        }

        public int search(String word) {  //word这个单词之前加入过几次
            if (word == null) {
                return 0;
            }
            char[] chs = word.toCharArray();
            TrieNode node = root;
            int index = 0;
            for (int i = 0; i < chs.length; i++) {
                index = chs[i] - 'a';
                if (node.nexts[index] == null) {
                    return 0;
                }
                node = node.nexts[index];
            }
            return node.end;
        }

        public int prefixNumber(String pre) {
            if (pre == null) {
                return 0;
            }
            char[] chs = pre.toCharArray();
            TrieNode node = root;
            int index = 0;
            for (int i = 0; i < chs.length; i++) {
                index = chs[i] - 'a';
                if (node.nexts[index] == null) {
                    return 0;
                }
                node = node.nexts[index];
            }
            return node.path;
        }
    }

    public static void main(String[] args) {
        Trie trie = new Trie();
        System.out.println(trie.search("zuo"));
        trie.insert("zuo");
        System.out.println(trie.search("zuo"));
        trie.delete("zuo");
        System.out.println(trie.search("zuo"));
        trie.insert("zuo");
        trie.insert("zuo");
        trie.delete("zuo");
        System.out.println(trie.search("zuo"));
        trie.delete("zuo");
        System.out.println(trie.search("zuo"));
        trie.insert("zuoa");
        trie.insert("zuoac");
        trie.insert("zuoab");
        trie.insert("zuoad");
        trie.delete("zuoa");
        System.out.println(trie.search("zuoa"));
        System.out.println(trie.prefixNumber("zuo"));
    }
}

贪心算法

在某一个标准下,优先考虑最满足标准的样本,最后考虑最不满足标准的样本,最终得到 一个答案的算法,叫作贪心算法。也就是说,不从整体最优上加以考虑,所做出的是在某种意义上的局部最优解。

局部最优-?->整体最优

贪心算法的在笔试时的解题套路

1, 实现一个不依靠贪心策略的解法X,可以用最暴力的尝试

2, 脑补出贪心策略A、贪心策略B、贪心策略C...

3, 用解法X和对数器,去验证每一个贪心策略,用实验的方式得知哪个贪心策略正确

4,不要去纠结贪心策略的证明

从头到尾展示最正统的贪心策略求解过程

例子:给定一个字符串类型的数组strs,找到一种拼接方式,使得把所有字符串拼起来之后形成的 字符串具有最小的字典序。证明贪心策略可能是件非常腌心的事情。平时当然推荐你搞清楚所有的来龙去脉,但是笔试 时用对数器的方式!

比较策略,要有传递性

import java.util.Arrays;
import java.util.Comparator;

public class LowestLexicography {

    public static class MyComparator implements Comparator<String> {
        @Override
        public int compare(String a, String b) {
            return (a + b).compareTo(b + a);
        }
    }

    public static String lowestString(String[] strs) {
        if (strs == null || strs.length == 0) {
            return "";
        }
        Arrays.sort(strs, new MyComparator());
        String res = "";
        for (int i = 0; i < strs.length; i++) {
            res += strs[i];
        }
        return res;
    }

    public static void main(String[] args) {
        String[] strs1 = { "jibw", "ji", "jp", "bw", "jibw" };
        System.out.println(lowestString(strs1));
        String[] strs2 = { "ba", "b" };
}

贪心策略在实现时,经常使用到的技巧:

1, 根据某标准建立一个比较器来排序

2, 根据某标准建立一个比较器来组成堆

一块金条切成两半,是需要花费和长度数值一样的铜板的。比如长度为20的金 条,不管切成长度多大的两半,都要花费20个铜板。一群人想整分整块金条,怎么分最省铜板?

例如,给定数组{10,20,30},代表一共三个人,整块金条长度为10+20+30=60。金条要分成10,20,30三个部分。如果先把长度60的金条分成10和50,花费60; 再把长度50的金条分成20和30,花费50; 一共花费110铜板。但是如果先把长度60的金条分成30和30,花费60;再把长度30金条分成10和20, 花费30; 一共花费90铜板。输入一个数组,返回分割的最小代价。

import java.util.Comparator;
import java.util.PriorityQueue;

public class LessMoneySplitGold {

    public static int lessMoney(int[] arr) {
        PriorityQueue<Integer> pQ = new PriorityQueue<>();
        for (int i = 0; i < arr.length; i++) {
            pQ.add(arr[i]);
        }
        int sum = 0;
        int cur = 0;
        while (pQ.size() > 1) {
            cur = pQ.poll() + pQ.poll();
            sum += cur;
            pQ.add(cur);
        }
        return sum;
    }

    public static class MinheapComparator implements Comparator<Integer> {

        @Override
        public int compare(Integer o1, Integer o2) {
            return o1 - o2; // < 0  o1 < o2  负数
        }

    }

    public static class MaxheapComparator implements Comparator<Integer> {

        @Override
        public int compare(Integer o1, Integer o2) {
            return o2 - o1; // <   o2 < o1
        }

    }

    public static void main(String[] args) {
        // solution
        int[] arr = { 6, 7, 8, 9 };
        System.out.println(lessMoney(arr));
        int[] arrForHeap = { 3, 5, 2, 7, 0, 1, 6, 4 };
        // min heap
        PriorityQueue<Integer> minQ1 = new PriorityQueue<>();
        for (int i = 0; i < arrForHeap.length; i++) {
            minQ1.add(arrForHeap[i]);
        }
        while (!minQ1.isEmpty()) {
            System.out.print(minQ1.poll() + " ");
        }
        System.out.println();
        // min heap use Comparator
        PriorityQueue<Integer> minQ2 = new PriorityQueue<>(new MinheapComparator());
        for (int i = 0; i < arrForHeap.length; i++) {
            minQ2.add(arrForHeap[i]);
        }
        while (!minQ2.isEmpty()) {
            System.out.print(minQ2.poll() + " ");
        }
        System.out.println();
        // max heap use Comparator
        PriorityQueue<Integer> maxQ = new PriorityQueue<>(new MaxheapComparator());
        for (int i = 0; i < arrForHeap.length; i++) {
            maxQ.add(arrForHeap[i]);
        }
        while (!maxQ.isEmpty()) {
            System.out.print(maxQ.poll() + " ");
        }
    }
}

一些项目要占用一个会议室宣讲,会议室不能同时容纳两个项目的宣讲。 给你每一个项目开始的时间和结束的时间(给你一个数组,里面是一个个具体 的项目),你来安排宣讲的日程,要求会议室进行的宣讲的场次最多。返回这个最多的宣讲场次。

import java.util.Arrays;
import java.util.Comparator;

    public static class Program {
        public int start;
        public int end;

        public Program(int start, int end) {
            this.start = start;
            this.end = end;
        }
    }

    public static class ProgramComparator implements Comparator<Program> { //比较器

        @Override
        public int compare(Program o1, Program o2) {
            return o1.end - o2.end;
        }
    }

    public static int bestArrange(Program[] programs, int start) {
        Arrays.sort(programs, new ProgramComparator());
        int result = 0;
        for (int i = 0; i < programs.length; i++) {  //遍历所有会议
            if (start <= programs[i].start) {
                result++;
                start = programs[i].end;
            }
        }
        return result;
    }
    public static void main(String[] args) {
    }
}

输入:正数数组costs 正数数组profits 正数k 正数m

含义:costs [i]表示i号项目的花费 profits [i]表示i号项目在扣除花费之后还能挣到的钱(利润)

k表示你只能串行的最多做k个项目 m表示你初始的资金

说明:你每做完一个项目,马上获得的收益,可以支持你去做下一个项目。

输出:你最后获得的最大钱数。

import java.util.Comparator;
import java.util.PriorityQueue;

public class IPO {
    public static class Node {
        public int p;
        public int c;

        public Node(int p, int c) {
            this.p = p;
            this.c = c;
        }
    }

    public static class MinCostComparator implements Comparator<Node> {

        @Override
        public int compare(Node o1, Node o2) {
            return o1.c - o2.c;
        }
    }

    public static class MaxProfitComparator implements Comparator<Node> {

        @Override
        public int compare(Node o1, Node o2) {
            return o2.p - o1.p;
        }
    }

    public static int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {
        Node[] nodes = new Node[Profits.length];
        for (int i = 0; i < Profits.length; i++) {
            nodes[i] = new Node(Profits[i], Capital[i]);
        }
        PriorityQueue<Node> minCostQ = new PriorityQueue<>(new MinCostComparator());
        PriorityQueue<Node> maxProfitQ = new PriorityQueue<>(new MaxProfitComparator());
        for (int i = 0; i < nodes.length; i++) {
            minCostQ.add(nodes[i]);
        }
        for (int i = 0; i < k; i++) {
            while (!minCostQ.isEmpty() && minCostQ.peek().c <= W) {
                maxProfitQ.add(minCostQ.poll());
            }
            if (maxProfitQ.isEmpty()) {
                return W;
            }
            W += maxProfitQ.poll().p;
        }
        return W;
    }
}

一个数据流中,随时可以取得中位数

import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;

public class MadianQuick {

    public static class MedianHolder {
        private PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(new MaxHeapComparator());
        private PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>(new MinHeapComparator());

        private void modifyTwoHeapsSize() {
            if (this.maxHeap.size() == this.minHeap.size() + 2) {
                this.minHeap.add(this.maxHeap.poll());
            }
            if (this.minHeap.size() == this.maxHeap.size() + 2) {
                this.maxHeap.add(this.minHeap.poll());
            }
        }

        public void addNumber(int num) {
            if (maxHeap.isEmpty() || num <= maxHeap.peek()) {
                maxHeap.add(num);
            } else {
                minHeap.add(num);
            }
            modifyTwoHeapsSize();
        }

        public Integer getMedian() {
            int maxHeapSize = this.maxHeap.size();
            int minHeapSize = this.minHeap.size();
            if (maxHeapSize + minHeapSize == 0) {
                return null;
            }
            Integer maxHeapHead = this.maxHeap.peek();
            Integer minHeapHead = this.minHeap.peek();
            if (((maxHeapSize + minHeapSize) & 1) == 0) {
                return (maxHeapHead + minHeapHead) / 2;
            }
            return maxHeapSize > minHeapSize ? maxHeapHead : minHeapHead;
        }

    }

    public static class MaxHeapComparator implements Comparator<Integer> {
        @Override
        public int compare(Integer o1, Integer o2) {
            if (o2 > o1) {
                return 1;
            } else {
                return -1;
            }
        }
    }

    public static class MinHeapComparator implements Comparator<Integer> {
        @Override
        public int compare(Integer o1, Integer o2) {
            if (o2 < o1) {
                return 1;
            } else {
                return -1;
            }
        }
    }

    // for test
    public static int[] getRandomArray(int maxLen, int maxValue) {
        int[] res = new int[(int) (Math.random() * maxLen) + 1];
        for (int i = 0; i != res.length; i++) {
            res[i] = (int) (Math.random() * maxValue);
        }
        return res;
    }

    // for test, this method is ineffective but absolutely right
    public static int getMedianOfArray(int[] arr) {
        int[] newArr = Arrays.copyOf(arr, arr.length);
        Arrays.sort(newArr);
        int mid = (newArr.length - 1) / 2;
        if ((newArr.length & 1) == 0) {
            return (newArr[mid] + newArr[mid + 1]) / 2;
        } else {
            return newArr[mid];
        }
    }

    public static void printArray(int[] arr) {
        for (int i = 0; i != arr.length; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        boolean err = false;
        int testTimes = 200000;
        for (int i = 0; i != testTimes; i++) {
            int len = 30;
            int maxValue = 1000;
            int[] arr = getRandomArray(len, maxValue);
            MedianHolder medianHold = new MedianHolder();
            for (int j = 0; j != arr.length; j++) {
                medianHold.addNumber(arr[j]);
            }
            if (medianHold.getMedian() != getMedianOfArray(arr)) {
                err = true;
                printArray(arr);
                break;
            }
        }
        System.out.println(err ? "Oops" : "beautiful ^_^");
    }
}

数据结构与算法——前缀树和贪心算法(1)

标签:字符串   --   拼接   答案   删除   tran   ring   als   c++   

原文地址:https://www.cnblogs.com/wwj99/p/12230745.html

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