@[TOC](acwing基础算法(20道题)【java版本】✔ ★)
一、快速排序
1. 快速排序例题
思想：小于x的放在左边，大于x的放在右边，等于x的放在左右两侧都可
关键：区间划分的时候，以i还是j为分界点
就看i和j停止时，所指向的数字，是大于x还是小于x
import java.util.*;
class Main {
static int N = (int)1e5+10;
static int[] nums = new int[N];
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
for (int i = 1; i <= n; i++) {
nums[i] = scanner.nextInt();
}
quickSort(nums,1,n);
for (int i = 1; i <= n; i++) {
System.out.print(nums[i] + " ");
}
}
private static void quickSort(int[] nums, int l, int r) {
if (l >= r) {
return;
}
int x = nums[l+r>>1];
int i = l-1;
int j = r+1;
while (i < j) {
do i++; while(nums[i] < x);
do j--; while(nums[j] > x);
if (i < j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
quickSort(nums,l,j);
quickSort(nums,j+1,r);
}
}
2. 第k个数( 快速选择 )
import java.util.*;
class Main {
static int N = (int)1e5+10;
static int[] nums = new int[N];
static int n,k;
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
n = scanner.nextInt();
k = scanner.nextInt();
for (int i = 1; i <= n; i++) {
nums[i] = scanner.nextInt();
}
System.out.print(quickSortK(1,n));
}
private static int quickSortK(int l, int r) {
if (l >= r) {
return nums[l];
}
int x = nums[l+r>>1];
int i = l-1;
int j = r+1;
while (i < j) {
do i++; while (nums[i] < x);
do j--; while (nums[j] > x);
if (i < j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
if (j >= k) {
return quickSortK(l,j);
} else {
return quickSortK(j+1,r);
}
}
}
二、归并排序
1. 归并排序模板题
import java.util.*;
class Main {
static int N = (int)1e5+10;
static int[] nums = new int[N];
static int[] temp = new int[N];
static int n;
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
n = scan.nextInt();
for (int i = 1; i <= n; i++) {
nums[i] = scan.nextInt();
}
mergeSort(1,n);
for (int i = 1; i <= n; i++) {
System.out.print(nums[i] + " ");
}
}
private static void mergeSort(int l, int r) {
if (l >= r) {
return;
}
int mid = l + r>>1;
mergeSort(l, mid);
mergeSort(mid+1, r);
int i = l, j = mid+1;
int k = l;
while (i <= mid && j <= r) {
if (nums[i] <= nums[j]) {
temp[k++] = nums[i++];
} else {
temp[k++] = nums[j++];
}
}
while (i <= mid) {
temp[k++] = nums[i++];
}
while (j <= r) {
temp[k++] = nums[j++];
}
k = l;
for (int q = l; q <= r; q++) {
nums[q] = temp[k++];
}
}
}
2. 逆序对的数量
import java.util.*;
public class Main {
public static long ans = 0;
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] nums = new int[n];
for (int i = 0; i < n; i++) {
nums[i] = scanner.nextInt();
}
numberOfReverseOrderPairs(0,n-1,nums);
System.out.print(ans);
}
public static void numberOfReverseOrderPairs(int l,int r,int[] nums){
if (l >= r) {
return;
}
int mid = (l+r)>>1;
int i = l,j = mid+1;
numberOfReverseOrderPairs(l,mid,nums);
numberOfReverseOrderPairs(mid+1,r,nums);
int[] temp = new int[r-l+1];
int k = 0;
while (i <= mid && j <= r) {
if (nums[i] <= nums[j]) {
temp[k++] = nums[i++];
} else {
temp[k++] = nums[j++];
ans += mid - i + 1;
}
}
while (i <= mid) {
temp[k++] = nums[i++];
}
while (j <= r) {
temp[k++] = nums[j++];
}
for (int q = l,p = 0; q <= r; q++) {
nums[q] = temp[p++];
}
}
}
三、二分
1. 数的范围
import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int[] nums = new int[n];
for (int i = 0; i < n; i++) {
nums[i] = scanner.nextInt();
}
for (int i = 0; i < m; i++) {
int x = scanner.nextInt();
int p = 0,q = 0;
int l = 0,r = n-1;
while (l < r) {
int mid = (l+r)>>1;
if (nums[mid] < x) {
l = mid+1;
} else {
r = mid;
}
}
p = r;
l = 0;
r = n-1;
while (l < r) {
int mid = (l+r+1)>>1;
if (nums[mid] <= x) {
l = mid;
} else {
r = mid - 1;
}
}
q = r;
if (nums[q] != x) {
System.out.println(-1 + " " + -1);
} else {
System.out.println(p + " " + q);
}
}
}
}
2. 数的三次方根
import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
double n = scanner.nextDouble();
double l = (double)-1e5;
double r = 1e5*1.0;
while (r - l > 1e-8) {
double mid = (l+r)/2;
if (mid * mid * mid <= n) {
l = mid;
} else {
r = mid;
}
}
System.out.printf("%.6f",l);
}
}
四、高精度
1. 高精度加法
import java.io.BufferedInputStream;
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
BigInteger a = scanner.nextBigInteger();
BigInteger b = scanner.nextBigInteger();
System.out.println(a.add(b));
scanner.close();
}
}
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String s1 = scanner.next();
String s2 = scanner.next();
List<Integer> A = new ArrayList<>(s1.length());
List<Integer> B = new ArrayList<>(s2.length());
for (int i = s1.length() - 1; i >= 0; i--) A.add(s1.charAt(i) - '0');
for (int i = s2.length() - 1; i >= 0; i--) B.add(s2.charAt(i) - '0');
List<Integer> C = add(A, B);
for (int i = C.size() - 1; i >= 0; i--) System.out.printf(C.get(i) + "");
}
private static List<Integer> add(List<Integer> A, List<Integer> B) {
List<Integer>C=new ArrayList<>(Math.max(A.size(),B.size())+2);
int t=0;
for(int i=0;i<A.size() || i<B.size();i++){
if(i<A.size())t+=A.get(i);
if(i<B.size())t+=B.get(i);
C.add(t%10);
t/=10;
}
if(t!=0) C.add(t);
return C;
}
}
2. 高精度减法
import java.io.*;
import java.math.BigInteger;
public class Main {
public static void main(String[] args) throws IOException{
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
BigInteger a = new BigInteger(reader.readLine());
BigInteger b = new BigInteger(reader.readLine());
System.out.println(a.subtract(b));
reader.close();
}
}
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class Main{
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
String s1 = scanner.next();
String s2 = scanner.next();
List<Integer> A = new ArrayList<>();
List<Integer> B = new ArrayList<>();
for(int i = s1.length() - 1;i >= 0;i --) A.add(s1.charAt(i) - '0');
for(int i = s2.length() - 1;i >= 0; i --) B.add(s2.charAt(i) - '0');
if(!cmp(A,B)){
System.out.print("-");
}
List<Integer> C = sub(A,B);
for(int i = C.size() - 1;i >= 0; i --){
System.out.print(C.get(i));
}
}
public static List<Integer> sub(List<Integer> A,List<Integer> B){
if(!cmp(A,B)){
return sub(B,A);
}
List<Integer> C = new ArrayList<>();
int t = 0;
for(int i = 0;i < A.size();i ++){
//这里应该是A.get(i) - B.get(i) - t ，因为可能B为零，所以需要判断一下是不是存在
t = A.get(i) - t;
if(i < B.size()) t -= B.get(i);
C.add((t + 10) % 10);
if(t < 0) t = 1;
else t = 0;
}
//删除指定下标下面的值
while(C.size() > 1 && C.get(C.size() - 1) == 0) C.remove(C.size() - 1);
return C;
}
public static boolean cmp(List<Integer> A,List<Integer> B){
if(A.size() != B.size()) return A.size() > B.size();
/* if(A.size() >= B.size()){
return true;
}else{
return false;
}
*/
for(int i = A.size() - 1;i >= 0;i --){
if(A.get(i) != B.get(i)) {
return A.get(i) > B.get(i);
}
}
return true;
}
}
3. 高精度乘法
二刷：
- 在草稿纸上演算一遍 运算过程，便知道 代码逻辑
原题链接
import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
String a = scanner.next();
int b = scanner.nextInt();
List<Integer> A = new ArrayList<>(a.length());
for (int i = a.length()-1; i >= 0; i--) A.add(a.charAt(i) - '0');
List<Integer> C = mul(A,b);
for (int i = C.size()-1; i >= 0; i--) System.out.print(C.get(i));
}
public static List<Integer> mul (List<Integer> A,int b) {
List<Integer> C = new ArrayList<>(A.size());
int t = 0;
for (int i = 0; i < A.size(); i++) {
t = t + A.get(i)*b;
C.add(t%10);
t /= 10;
}
while (t != 0) {
C.add(t%10);
t /= 10;
}
while (C.size() > 1 && C.get(C.size()-1) == 0) {
C.remove(C.size() - 1);
}
return C;
}
}
4. 高精度除法
import java.util.*;
class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String a = scan.nextLine();
int b = scan.nextInt();
ArrayList<Integer> A = new ArrayList<Integer>();
for (int i = 0; i < a.length(); i++) {
A.add(a.charAt(i)-'0');
}
ArrayList<Integer> C = div(A, b);
}
private static ArrayList<Integer> div(ArrayList<Integer> A, int b) {
ArrayList<Integer> C = new ArrayList<Integer>();
int t = 0;
for (int i = 0; i < A.size(); i++) {
t = t * 10 + A.get(i);
if (t / b > 0) {
C.add(t / b);
t = t % b;
} else {
C.add(0);
}
}
while (C.size() > 1 && C.get(0) == 0) C.remove(0);
for (int i = 0; i < C.size(); i++) {
System.out.print(C.get(i));
}
System.out.println();
System.out.print(t);
return C;
}
}
五、前缀和S与差分a
1. 前缀和
import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int[] a = new int[n+1];
int[] s = new int[n+1];
for (int i = 1; i <= n; i++) { a[i] = scanner.nextInt(); }
s[1] = a[1];
for (int i = 2; i <= n; i++) {
s[i] = a[i] + s[i-1];
}
while (m-- != 0) {
int l = scanner.nextInt();
int r = scanner.nextInt();
System.out.println(s[r]-s[l-1]);
}
}
}
2. 子矩阵的和
import java.util.*;
class Main {
static int N = 1010;
static int[][] nums = new int[N][N];
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int l = scan.nextInt();
int r = scan.nextInt();
int q = scan.nextInt();
for (int i = 1; i <= l; i++) {
for (int j = 1; j <= r; j++) {
nums[i][j] = scan.nextInt();
nums[i][j] = nums[i][j] + nums[i][j-1] + nums[i-1][j] - nums[i-1][j-1];
}
}
for (int i = 0; i < q; i++) {
int x1 = scan.nextInt();
int y1 = scan.nextInt();
int x2 = scan.nextInt();
int y2 = scan.nextInt();
System.out.println(nums[x2][y2] - nums[x2][y1-1] - nums[x1-1][y2] + nums[x1-1][y1-1]);
}
}
}
3. 差分
- 由差分求s时，是要有数据连续性的，前面的改变了，要保证对应后面的也跟着
import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int[] nums = new int[n+2];
for (int i = 1; i <= n; i++) {
nums[i] = scanner.nextInt();
}
for (int i = n; i >= 1; i--) {
nums[i] = nums[i]-nums[i-1];
}
for (int i = 0; i < m; i++) {
int l = scanner.nextInt();
int r = scanner.nextInt();
int c = scanner.nextInt();
nums[l] += c;
nums[r+1] -= c;
}
for (int i = 1; i <= n; i++) {
nums[i] += nums[i-1];
System.out.printf("%d ",nums[i]);
}
}
}
4. 差分矩阵
import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int q = scanner.nextInt();
int[][] splits = new int[n+2][m+2];
int[][] sum = new int[n+2][m+2];
for (int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
sum[i][j] = scanner.nextInt();
splits[i][j] = sum[i][j] - sum[i-1][j] - sum[i][j-1] + sum[i-1][j-1];
}
}
for (int i = 0; i < q; i++) {
int x1 = scanner.nextInt();
int y1 = scanner.nextInt();
int x2 = scanner.nextInt();
int y2 = scanner.nextInt();
int c = scanner.nextInt();
splits[x1][y1] += c;
splits[x1][y2+1] -= c;
splits[x2+1][y1] -= c;
splits[x2+1][y2+1] += c;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
sum[i][j] = splits[i][j] + sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];
System.out.print(sum[i][j] + " ");
}
System.out.println();
}
}
}
六、双指针
1. 最长连续不重复子序列
import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int[] a = new int[100010];
int[] s = new int[100010];
for(int i = 0 ; i < n ; i ++ ){
a[i] = scan.nextInt();
}
int res = 0;
for(int i = 0 , j = 0 ;i < n ; i ++ ){
//比如一开始S[2]是0；然后你的a[1] = 2;那么s[2] = 1;
//然后如果a[2] = 2 ;那么第二次出现所以s[2] = 2;这样来证明是不是出现两次
s[a[i]] ++ ;
while(j < i && s[a[i]] > 1){
//一开始j是跟i在同个位置，i在移动，j原地不动，只要上面出现两次，j开始移动
//j移动到i的位置，下面的代码就是i走过的路，让s[i] 数组里面加1的位置全部减1，就变回0；所以就继续往下统计长度
s[a[j]] -- ;
j++;
}
//i-j+1是统计长度的公式；
res = Math.max(res, i-j+1);
}
System.out.println(res);
}
}
//这里填你的代码^^
//注意代码要放在两组三个点之间，才可以正确显示代码高亮哦~
2. 数组元素的目标和
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int x = scanner.nextInt();
int[] A = new int[n+1];
int[] B = new int[m+1];
for (int i = 0; i < n; i++) A[i] = scanner.nextInt();
for (int i = 0; i < m; i++) B[i] = scanner.nextInt();
for (int i = 0,j = m-1; i < n && j >= 0;) {
if (A[i] + B[j] < x) {
i++;
} else if (A[i] + B[j] == x) {
System.out.print(i + " " + j);
break;
} else {
j--;
}
}
}
}
3. 判断子序列
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int[] a = new int[n+1];
int[] b = new int[m+1];
for (int i = 0; i < n; i++) a[i] = scanner.nextInt();
for (int j = 0; j < m; j++) b[j] = scanner.nextInt();
boolean flag = false;
for (int i = 0,j = 0; j < m; j++) {
if (a[i] == b[j]) {
i++;
if (i == n) {
System.out.print("Yes");
flag = true;
break;
}
}
}
if (flag == false) {
System.out.print("No");
}
}
}
七、二进制
1. 位运算算法
返回n的最后一位1：lowbit(n) = n & -n
一共有多少1 ： while n = n ^（n & -n）或者 n -= n & -n
原题链接
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] nums = new int[n];
for (int i = 0; i < n; i++) {
nums[i] = scanner.nextInt();
System.out.print(numberOfOneInBinary(nums[i]) + " ");
}
}
public static int numberOfOneInBinary(int num) {
int cnt = 0;
while (num != 0) {
num -= (num & -num);
cnt++;
}
return cnt;
}
}
八、离散化
1. 区间和
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
pair[] add = new pair[n+2];
pair[] query = new pair[m];
List<Integer> arrayMap = new ArrayList<>(n+m*2);
int[] sum = new int[n+m*2];
for (int i = 0; i < n; i++) {
add[i] = new pair();
add[i].l = scanner.nextInt();
add[i].r = scanner.nextInt();
arrayMap.add(add[i].l);
}
for (int i = 0; i < m; i++) {
query[i] = new pair();
query[i].l = scanner.nextInt();
query[i].r = scanner.nextInt();
arrayMap.add(query[i].l);
arrayMap.add(query[i].r);
}
arrayMap = new ArrayList<>(new HashSet<>(arrayMap));
Collections.sort(arrayMap);
for (int i = 0; i < n; i++) {
sum[arrayMapIndexOf(add[i].l,arrayMap)] += add[i].r;
}
for (int i = 0; i < arrayMap.size(); i++) {
if(i != 0) {
sum[i] += sum[i-1];
}
}
for (int i = 0; i < query.length; i++) {
int l = arrayMapIndexOf(query[i].l,arrayMap);
int r = arrayMapIndexOf(query[i].r,arrayMap);
if (l == 0) {
System.out.print(sum[r] + "\n");
} else {
System.out.print(sum[r]-sum[l-1] + "\n");
}
}
}
static class pair {
int l;
int r;
}
private static int arrayMapIndexOf(int k,List<Integer> arrayMap) {
int l = 0,r = arrayMap.size()-1;
while (l < r) {
int mid = (l+r+1) >> 1;
if (arrayMap.get(mid) <= k) {
l = mid;
} else {
r = mid - 1;
}
}
return r;
}
}
九、区间合并
1. 区间合并
贪心做法
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
pair[] range = new pair[n];
for (int i = 0; i < n; i++) {
range[i] = new pair();
range[i].l = scanner.nextInt();
range[i].r = scanner.nextInt();
}
Arrays.sort(range,new Comparator<pair>(){
@Override
public int compare(pair o1,pair o2) {
if (o1.l == o2.l) {
return o1.r - o2.r;
} else {
return o1.l - o2.l;
}
}
});
int st = (int)-2e9-1;
int ed = (int)-2e9-1;
int cnt = 0;
for (int i = 0; i < n; i++) {
if (ed < range[i].l) {
st = range[i].l;
ed = range[i].r;
cnt++;
} else {
ed = Math.max(ed,range[i].r);
}
}
System.out.print(cnt);
}
static class pair {
int l;
int r;
}
}