标签:字符串 -- 拼接 答案 删除 tran ring als c++
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"));
}
}
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
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
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" };
}
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;
}
}
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 ^_^");
}
}
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