Introduction prefix tree
What is the prefix tree? How to generate a prefix tree?
Examples: a string type array arrl, another string array of type arr2. arr2 in which characters are arr 1 appear in? Please print. arr2 in which characters in a string as a prefix arr occur? Please print. arr2 in which characters are in a string prefix as arr1 occur? Please print arr2 the maximum number of prefixes appear.
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"));
}
}how are you
In a certain standard, priority sample standard of the most satisfying, most final consideration does not meet the standard sample, finally get an answer algorithm, called greedy algorithm. In other words, not be considered as a whole the best, the optimal solution is made in the local sense.
Local optimization -? -> Global Optimization
Greedy algorithm solving routine at the time of the written test
1, to achieve a greedy strategy does not rely on the solution X, you can use the most violent attempt
2, the brain make the greedy strategy A, greedy strategy B, C ... greedy strategy
3, with solution X and logarithmic, a strategy to verify each greedy, greedy strategy which experimentally that the correct way
4, not to tangle proof greedy strategy
Show from start to finish most orthodox solution process greedy strategy
Examples: Given a string type array of STRs, splicing find a way that the string formed after having put together all the strings lexicographically smallest. Greedy strategy may prove very marinade is a heart thing. Of course, usually recommend that you find out all the ins and outs, but the way the log is written with time!
Comparing policies, there must be transitive
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" };
}Greedy strategy when implemented, are frequently used skills:
1, the establishment of a comparator according to a sorting criteria
2, the establishment of a comparator according to a standard group of piled
One
A bullion cut in half, and takes a value the same as the length of the copper plate. For example, a length of 20 gold bars, cut into two halves regardless of how much length, must spend 20 copper. A group of people who want to divide the entire block of gold bullion, how to divide the province most copper?
For example, given the array {10, 20}, representing a total of three, the entire length of bars 10 + 20 + 30 = 60. Bars 10, 20 to be divided into three parts. If the length of the first bars 10 and 60 into 50, take 60; 50 and then into longitudinal bars 20 and 30, it takes 50; 110 takes a total of copper. However, if the length of the first bars 30 and 60 into 30, take 60; 30 then the length of the bars 10 and 20 into spent 30; 90 spent a total of copper. An input array, divided returns minimum cost.
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() + " ");
}
}
}two
Some projects take up a conference room preach, preach meeting room can not accommodate both projects at the same time. Each time you start a project and end time (to give you an array, which is a concrete project), you preach to arrange the schedule, conference room requires the screening of up to preach. Returns the most preaching sessions.
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) {
}
}three
Input: Array costs positive positive positive profits array numbers k m n
Meaning: costs [i] represents the cost profits [i] i i represents the number of projects number project after deducting the cost but also earn money (profit)
k k means that you can only do projects that you m initial funding of up to serial
Description: Every time you finish a project and immediately gains that can support you to do the next project.
Output: maximum money you finally get the number.
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;
}
}A data stream, you can always get a median
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 ^_^");
}
}