AcWing 算法基础课二数据结构链表栈队列并查集哈希表

单链表.

AcWing. 826.单链表

import java.util.Scanner;
public class Main{static int[] e = new int[100010];//结点i的值static int[] ne = new int[100010];//结点i的next指针static int idx,head;//head是头结点，idx存当前已经用到了哪个点public static void init(){idx = 0;head = -1;}//H向链表头插入一个数x；public static void add_head(int x){e[idx] = x;ne[idx] = head;head = idx++;} //I在第k位数后面插入一个数xpublic static void add(int k,int x){e[idx] = x;ne[idx] = ne[k];ne[k] = idx++;}//D删除第k个数后面得数public static void remove(int k){ne[k] = ne[ne[k]];}public static void main(String[] args){Scanner scan = new Scanner(System.in);int m = scan.nextInt();init();while(m -- > 0){//因为java中没有输入一个字符，所以用字符串转字符String s = scan.next();char op = s.charAt(0);if(op == 'H'){int x = scan.nextInt();add_head(x);}else if(op == 'D'){int k = scan.nextInt();if(k == 0) head = ne[head];else remove(k-1);}else {int k = scan.nextInt();int x = scan.nextInt();add(k-1,x);}}for(int i = head;i != -1;i = ne[i] ){System.out.print(e[i] +  " ");}}
}

双链表

AcWing 827.双链表

import java.util.Scanner;
public class Main{static int N = 100010;static int[] e = new int[N];//元素static int[] r = new int[N];//右指针static int[] l = new int[N];//左指针static int idx;//删除第k位插入的数public static void remove(int k){r[l[k]] = r[k];l[r[k]] = l[k];}//这是在第k位数后面插入一个数xpublic static void add_all(int k,int x){e[idx] = x;r[idx] = r[k];l[idx] = k;l[r[k]] = idx;r[k] = idx++;}public static void main(String[] args){Scanner scan = new Scanner(System.in);int m = scan.nextInt();r[0] = 1;l[1] = 0;idx = 2;while(m -- > 0){String s = scan.next();char op = s.charAt(0);if(op == 'L'){int x = scan.nextInt();add_all(0,x);}else if(op == 'R'){int x = scan.nextInt();add_all(l[1],x);}else if(op == 'D'){int k = scan.nextInt();remove(k + 1);}else if(s.equals("IR")){int k  = scan.nextInt();int x = scan.nextInt();add_all(k + 1,x);}else {int k = scan.nextInt();int x = scan.nextInt();add_all(l[k+1],x);}}for(int i = r[0];i != 1; i= r[i]){System.out.print(e[i]  + " ");}}
}

栈

AcWing 828.模拟栈

import java.util.Scanner;
public class Main{public static void main(String[] args){Scanner scan = new Scanner(System.in);int m = scan.nextInt();int[] stl = new int[100010];int tt = 0;while(m -- > 0){String s = scan.next();if(s.equals("push")){int x= scan.nextInt();stl[++tt] = x;}else if(s.equals("pop")){tt--;}else if(s.equals("empty")){if(tt > 0){System.out.println("NO");}else System.out.println("YES");}else{System.out.println(stl[tt]);}}}
}

AcWing 3302.表达式求值

import java.util.*;
public class Main{public static void main(String[] args){Scanner scan = new Scanner(System.in);//以字符串形式输入表达式String s = scan.next();//map来添加运算符号进去，定义优先级Map<Character,Integer> map = new HashMap<>();map.put('+',1);map.put('-',1);map.put('*',2);map.put('/',2);Stack<Character> op = new Stack<>();//存运算符号Stack<Integer> num = new Stack<>();//存数字for(int i = 0 ; i < s.length(); i ++ ){char c = s.charAt(i);//判断c字符是不是数字if(Character.isDigit(c)){int x = 0,j = i;//数字可能会是多位数，while(j < s.length() && Character.isDigit(s.charAt(j))){x = x * 10 + s.charAt(j) - '0';j++;}num.push(x);//将数字x存入数字栈栈顶i = j - 1;//重新赋值i}else if(c == '('){op.push(c); // 将左括号存入字符栈栈顶}else if(c == ')'){//如果栈顶不等于左括号，一直计算下去；while(op.peek() != '('){eval(op,num);}op.pop(); // 将左括号弹出栈顶}else { //如果是正常字符while(!op.empty() && op.peek() != '(' && map.get(op.peek()) >= map.get(c)){eval(op,num);}op.push(c);}}while(!op.empty()) eval(op,num);System.out.println(num.peek());}public static void eval(Stack<Character> op,Stack<Integer> num){int b = num.pop();int a = num.pop();char c = op.pop();if(c == '+'){num.push(a+b);}else if(c == '-'){num.push(a-b);}else if(c == '*'){num.push(a*b);}else {num.push(a/b);}}
}

队列

AcWing 829. 模拟队列

import java.util.Scanner;
public class Main{public static void main(String[] args){Scanner scan = new Scanner(System.in);int m = scan.nextInt();//队列是在tt队尾插入元素，队头hh弹出元素int[] dl = new int[100010];int hh = 0;int tt = -1;while(m -- > 0){String s = scan.next();if(s.equals("push")){int x = scan.nextInt();dl[++tt] =  x;}else if(s.equals("pop")){hh++;}else if(s.equals("empty")){if(hh <= tt) System.out.println("NO");else System.out.println("YES");}else {System.out.println(dl[hh]);}}}
}

单调栈

AcWing 830.单调栈

import java.util.*;
public class Main{public static void main(String[] args){Scanner scan = new Scanner(System.in);int n = scan.nextInt();int[] stk  = new int[100010];int tt = 0;for(int i = 0 ; i < n ;  i++ ){int x = scan.nextInt();//如果栈非空 ，栈顶元素>= x，那么说明栈顶这个数不满足需求，//因为比较的是左边第一个最近最小的数，所以就把stk[tt]弹出了；while(tt!=0 && stk[tt] >= x){tt--;}//如果弹出操作完了之后，栈不是空的，就输出栈顶元素，if(tt != 0) System.out.print(stk[tt]+" ");//否则就是栈是空的，输出-1else System.out.print("-1" + " ");//最后将x插入栈顶；stk[++tt] = x;}}
}

单调队列

AcWing 154.滑动窗口

import java.io.*;
public class Main{static int N = 1000010;static int n,k;//数组长度 和 滑动窗口长度static int hh,tt;//队头队尾static int[] q = new int[N];//q是队列 存的是 下标static int[] a = new int[N];public static void main(String[] args)throws IOException{BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));PrintWriter wt = new PrintWriter(new OutputStreamWriter(System.out));String[] st = bf.readLine().split(" ");n = Integer.parseInt(st[0]);k = Integer.parseInt(st[1]);String[] str = bf.readLine().split(" ");for(int i = 0 ; i < n ; i ++ ) a[i] = Integer.parseInt(str[i]);hh = 0;tt = - 1;//队头队尾 感觉像双指针for(int i = 0 ; i < n ; i ++ ){//i是 滑动窗口的 终点//队列非空 hh <= tt且 队头在起点之前 就出栈if(hh <= tt && q[hh] < i - k + 1) hh ++;//起点:i-k+1//这个while 就是 控制 这个队列的单调性递增！//非空 & 队尾元素 >= 新加进来的元素,就将队尾删掉，因 队头存的是窗口中最小的while(hh <= tt && a[q[tt]] >= a[i]) tt --;  //求最小值所以保留最小值q[++ tt] = i;//当前 位置插到 队尾//不足k 不用判断 if(i >= k - 1) wt.print(a[q[hh]] + " ");}wt.println();hh = 0;tt = -1;for(int i = 0 ; i < n ; i ++){if(hh <= tt && q[hh] < i - k + 1) hh ++;while(hh <= tt && a[q[tt]] <= a[i]) tt --;//求最大值所以保留最大值q[++ tt] = i;if(i >= k - 1) wt.print(a[q[hh]] + " ");}wt.flush();//记得刷新流}
}

KMP

AcWing 831. KMP字符串

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Scanner;
public class Main{static int N = 100010;static int ne[] = new int[N];//kmp核心数组next[];public static void main(String[] args) throws IOException {BufferedReader br = new BufferedReader(new InputStreamReader(System.in));BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));Integer n = Integer.parseInt(br.readLine());//输入p字符的长度String s1 = " " + br.readLine();//输入p字符串Integer m = Integer.parseInt(br.readLine());String s2 = " " + br.readLine();char[] a1 = s1.toCharArray();//创建p数组 存字符char[] a2 = s2.toCharArray();/*** ne[]：存储一个字符串以每个位置为结尾的‘可匹配最长前后缀’的长度。* next[i]=j 等价 p[1,j]=p[i-j+1,i]* 构建ne[]数组：*              1,初始化ne[1] = 0,i从2开始。当第一个数时就是0所以不用参与。默认是0；*              2,若匹配，s[i]=s[j+1]说明1~j+1是i的可匹配最长后缀,ne[i] = ++j;*              3,若不匹配，则从j的最长前缀位置+1的位置继续与i比较*              (因为i-1和j拥有相同的最长前后缀，我们拿j的前缀去对齐i-1的后缀),*              即令j = ne[j],继续比较j+1与i，若匹配转->>2*              4,若一直得不到匹配j最终会降到0，也就是i的‘可匹配最长前后缀’的长度*              要从零开始重新计算*/for(int i = 2,j = 0;i <= n ;i++) {//这里判断从j是不是>=0，如果< 0，表示还没有前缀和后缀相等while(j!=0&&a1[i]!=a1[j+1]) j = ne[j]; //不相等就让j等于上一个数的next[];if(a1[i]==a1[j+1]) j++;//匹配 有一个相等，前缀和后缀相等长度为1.ne[i] = j;}/*** 匹配两个字符串：*      1，从i=1的位置开始逐个匹配，利用ne[]数组减少比较次数*      2，若i与j+1的位置不匹配（已知1~j匹配i-j~i-1)，*      j跳回ne[j]继续比较(因为1~j匹配i-j~i-1,所以1~ne[j]也能匹配到i-ne[j]~i-1)*      3，若匹配则j++,直到j==n能确定匹配成功*      4，成功后依然j = ne[j],就是把这次成功当成失败，继续匹配下一个位置*/for(int i = 1,j = 0; i <= m;i++) {while(j!=0&&a2[i]!=a1[j+1]) j = ne[j];if(a2[i]==a1[j+1]) j++;if(j==n) {//相等的数= 短的数组的长度，就说明该输出了j = ne[j];//输出之后，要继续往下面遍历对比，所以让j=上一个数的缀长度，bw.write(i-n+" ");}}/*** 时间复杂度：*      因为：j最多加m次，再加之前j每次都会减少且最少减一，j>0*      所以：while循环最多执行m次,若大于m次，j<0矛盾*      最终答案：O(2m)*///所有write下的内容，会先存在writers中，当启用flush以后，会输出存在其中的内容。//如果没有调用flush，则不会将writer中的内容进行输出。bw.flush();}
}

Trie：高效地存储和查找字符串集合的数据结构

AcWing 835. Trie字符串统计

import java.util.Scanner;
public class Main{static int N = 100010,idx = 0;//idx是当前用到的下标static int[][] song = new int[N][26];//每个点的儿子集合最多26个字母儿子static int[] cnt  = new int[N];//存当前结尾的单词有多少个static char[] str = new char[N];//插入字符串public static  void insert(char[] str){int p = 0; //下标0表示头结点，根节点for(int i = 0 ; i < str.length; i ++ ){// 将字符串每个字符a-z 映射成数字；0-25int u = str[i] - 'a'; //如果这个儿子分支没有字符，说明这条分支还没有这个字符插入过//就创建出来 把【idx】下标赋值上去，作为每个分支的专属坐标if(song[p][u] == 0) song[p][u] = ++idx;//然后将p往下前进一层p = song[p][u];}//最后停在那一层的那个数字就做标记，说明这是要找的字符串的结尾cnt[p]++;}public static int query(char[] str){int p = 0;//从根节点开始，下标是0表示根节点，头结点for(int i = 0 ; i < str.length; i ++){int u = str[i] - 'a'; // 将字符串每个字符都转化成数字0-25//如果这个点上没有标记，就说明没有存入过这个字符，所以返回0if(song[p][u] == 0) return 0;//如果这个点上面能寻找到这个字符，就让他往下一层继续寻找；p = song[p][u];}//最后查找完之后输出最后一个做标记的点为下标的cnt数组的值。return cnt[p];}public static void main(String[] args){Scanner scan = new Scanner(System.in);int n = scan.nextInt();String sss = scan.nextLine();while(n -- > 0){String s = scan.nextLine();String[] st = s.split(" ");String s1 = st[0];String s2 = st[1];if(s1.equals("I")){insert(s2.toCharArray());}else{System.out.println(query(s2.toCharArray()));}}}
}

AcWing 143. 最大异或对

import java.io.*;
public class Main{static int N = 3100010,idx = 0;static int[][] song = new int[N][2];//插入public static void add(int x){int p = 0;//从头结点开始for(int i = 30 ; i >= 0 ; i -- ){ //从第31位开始遍历int u = x >> i & 1;//取出第i位的二进制数//判断这一层是空的，就创建，然后赋值下标if(song[p][u] == 0) song[p][u] = ++idx;//我觉得这个idx不是代表1或0，二维数组的u才代表这一条路的节点数字p = song[p][u];//然后让往下前进一层}}//查询public static int query(int x){int p = 0,res = 0;//从根节点0开始；res存的是这条路径表示的数for(int i = 30; i>= 0 ; i --){int u = x >> i & 1; //取出第i位的二进制数
//如果该节点的u是0，则判断一下在这一层有没有跟他相反的0，这样方便找最大异或值if(song[p][1-u] != 0){ res += (1 << i);//res就将该二进制位对应异或之后的最优解1每一位顺次加起来。因为是异或相反数就是1，这是最优解p = song[p][1-u];//然后往最优解那边前进一层。}else{//否则就没有 最优解的0匹配1,1匹配0，所以就异或之后的值是0//res += (0 << i);因为是0所以可以省略,p = song[p][u];//然后让他往不优解那边前进一层。}}return res;//最后返回异或之后的最大值res}public static void main(String[] args)throws IOException{BufferedReader re = new BufferedReader(new InputStreamReader(System.in));BufferedWriter wt = new BufferedWriter(new OutputStreamWriter(System.out));int n = Integer.parseInt(re.readLine());String[] s = re.readLine().split(" ");for(int i = 0 ; i < n ; i ++ ){add(Integer.parseInt(s[i]));}int res  = 0;for(int i = 0 ; i < n ; i ++ ){//因为输入的是字符串所以需要转成整形。然后每一次比较res的值谁大，然后将最大值重新赋值给resres = Math.max(res,query(Integer.parseInt(s[i])));}wt.write(res +" ");//最后输出res，因为快输出输出的是字符串，所以需要在后面加上“ ”；wt.close();}
}

并查集

AcWing 836.合并集合

import java.io.*;public class Main {private static int[] p = new int[(int) 10e5];private static int find(int x) {if (p[x] != x) p[x] = find(p[x]);return p[x];}public static void main(String[] args) throws IOException {BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));String[] s = reader.readLine().split(" ");int n = Integer.parseInt(s[0]), m = Integer.parseInt(s[1]);for (int i = 0; i < n; i++) {p[i] = i;}while (m-- > 0) {s = reader.readLine().split(" ");if ("Q".equals(s[0])) {int x = Integer.parseInt(s[1]);int y = Integer.parseInt(s[2]);writer.write(find(x) == find(y) ? "Yes\n" : "No\n");} else {int x = Integer.parseInt(s[1]);int y = Integer.parseInt(s[2]);p[find(x)] = find(y);}}writer.flush();}
}

AcWing 837. 连通块中点的数量

import java.io.*;public class Main {private static int[] p = new int[(int) 10e5];private static int find(int x) {if (p[x] != x) p[x] = find(p[x]);return p[x];}public static void main(String[] args) throws IOException {BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));String[] s = reader.readLine().split(" ");int n = Integer.parseInt(s[0]), m = Integer.parseInt(s[1]);int[] size = new int[n];for (int i = 0; i < n; i++) {p[i] = i;size[i]=1;}while (m-- > 0) {s = reader.readLine().split(" ");if ("Q1".equals(s[0])) {//是否在同一块内int x = Integer.parseInt(s[1]);int y = Integer.parseInt(s[2]);writer.write(find(x) == find(y) ? "Yes\n" : "No\n");} else if ("Q2".equals(s[0])) {//同一块内有多少节点int x = Integer.parseInt(s[1]);writer.write(size[find(x)]+"\n");} else {int x = Integer.parseInt(s[1]);int y = Integer.parseInt(s[2]);if(find(x) == find(y) ) continue;size[find(y)]+=size[find(x)];//别加反了p[find(x)] = find(y);}}writer.flush();}
}

AcWing 240.食物链

堆

AcWing 838.堆排序

import java.io.*;
public class Main{static int[] h=new int[100010];static int size;public static void main(String[] args) throws IOException {BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));String[] s = reader.readLine().split(" ");int n = Integer.parseInt(s[0]), m = Integer.parseInt(s[1]);s = reader.readLine().split(" ");for (int i = 1; i <= n; i++) {h[i]=Integer.parseInt(s[i-1]);//  writer.write(h[i]+"\n");} size=n;for(int i=n/2;i>0;i--){down(i);}while(m-->0) {writer.write(h[1]+" ");h[1]=h[size];size--;down(1);} writer.flush();}public static void down(int u){int t=u;if(u*2<=size&&h[u*2]<h[t]) t=u*2;if(u*2+1<=size&&h[u*2+1]<h[t]) t=u*2+1;if(t!=u){int tem=h[t];h[t]=h[u];h[u]=tem;down(t);}}
}

AcWing 839.模拟堆

哈希表

数据范围是1e9，所以需要先% 再加再%

AcWing 840.模拟散列表

拉链法(数组+单链表)

import java.io.*;
import java.util.Arrays;
public class Main{static int N=100003,idx=0;static int[] h=new int[N];static int[] e=new int[N];static int[] ne=new int[N];public static void main(String[] args) throws IOException {BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));String[] s = reader.readLine().split(" ");int n = Integer.parseInt(s[0]);//数组赋初始值Arrays.fill(h, -1);while(n-->0){s = reader.readLine().split(" ");int x=Integer.parseInt(s[1]);if("I".equals(s[0])) {//插入insert(x);}else{if( find(x) )writer.write("Yes\n");else  writer.write("No\n");}} writer.flush();}public static void insert (int x){int k=(x%N+N)%N;e[idx]=x;ne[idx]=h[k];h[k]=idx++;}public static boolean find (int x){int k=(x%N+N)%N;for(int i=h[k];i!=-1;i=ne[i]){if(e[i]==x){return true; }}return false;}
}

开放寻址法（蹲坑法）直接一个数组模拟蹲坑

AcWing 841.字符串哈希：快速判断两个字符串是否相等

unsigned long long 是2^64次代替了%Q,溢出时相当于自动取模

import java.util.Scanner ;
public class Main{//开long型数组，本来是 前缀hash求完之后 % 2的64次方来避免 冲突，可能超过我们给的数组大小static int N = 100010,P = 131;//p是进制数，经验值static long[] h = new long[N];// 存放hash前缀值static long[] p = new long[N];// 存放p的n次方public static long get(int l,int r){//求l-r区间的hash值，要用h[r] - h[l-1] 位数不同需要让h[l-1]向左边移到跟h[r]对齐//如求1234的3-4区间位，1234 - 12,12左移然后就让12*10^(4-3+1)=1200, 相减= 34//本题是p进制，需要将上行公式中的10换成p就行//h[0] = 0//h[1] = h[i-1] * P + str[1] = 0*P+a = a//h[2] = a * P + b//h[3] = (a*P+b)*P+c = a*p[2]+b*P+c//h[4] = (a*p[2]+b*P+c)*P+d = a*p[3]+b*p[2]+c*P+d//比如abcd求3-4区间位，就是让h[d]-h[b]，h[b]位数不用需要向左移对齐h[d],//h[2]*P^(4-3+1)=(a*P+b)*P^2 = a*P^3+b*P^2//然后 h[d] - h[b] 求34区间值,(a*p[3]+b*p[2]+c*P+d) - (a*P^3+b*P^2) = c*P+dreturn h[r] - h[l-1]*p[r-l+1];}public static void main(String[] args){Scanner scan = new Scanner(System.in);int n = scan.nextInt();int m = scan.nextInt();String s = scan.next();p[0] = 1;//p的0次方=1for(int i = 1 ; i <= n ; i++ ){p[i] = p[i-1] * P;//每一个下标对应P的多少次方//预处理公式 前缀哈希的值，P进制，中间*P    h[i] = h[i-1] * P + s.charAt(i-1);}while(m -- > 0){int l1 = scan.nextInt();int r1 = scan.nextInt();int l2 = scan.nextInt();int r2 = scan.nextInt();//判断两个区间是不是相同,用get的方法返回值一样说明区间的hash值是一样的if(get(l1,r1) == get(l2,r2)) System.out.println("Yes");else System.out.println("No");}}
}