✔ ★算法基础笔记（Acwing）（一）—— 基础算法(20道题)【java版本】

一、快速排序

1. 快速排序例题

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();
        }
        quickSort(0,n-1,nums);
        for (int i=0; i<n; i++) {
            System.out.print(nums[i] + " ");
        }
    }
    
    public static void quickSort (int l,int r,int[] nums) {
        if(l>=r) {
            return;
        }
        int x = nums[(l+r)/2];
        int i = l - 1,j = r + 1;
        while (i < j) {
            while (nums[++i] < x);
            while (nums[--j] > x);
            if (i < j) {
                int t = nums[i];
                nums[i] = nums[j];
                nums[j] = t;
            }
        }
        quickSort(l,j,nums);
        quickSort(j+1,r,nums);
    }
}

2. 第k个数( 快速选择 ) ✔ ✔1.31

原题链接

import java.util.*;

public class Main {
    public static int k;
    public static void main (String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        k = scanner.nextInt();
        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            nums[i] = scanner.nextInt();
        }
        System.out.print(quickSortTheK_thNumber(0,n-1,nums));
    }
    
    public static int quickSortTheK_thNumber (int l,int r,int[] nums) {
        if (l >= r) {
            return nums[r];
        }
        int x = nums[(l+r)>>1];
        int i = l - 1, j = r + 1;
        while (i < j) {
            while (nums[++i] < x);
            while (nums[--j] > x);
            if (i < j) {
                int t = nums[i];
                nums[i] = nums[j];
                nums[j] = t;
            }
        }
        if (j < k-1) {
            return quickSortTheK_thNumber(j+1,r,nums);
        } else {
            return quickSortTheK_thNumber(l,j,nums);
        }
    }
}

★快排二刷总结( 4点 )

1. if (l >= r) return;
2. 没有等于号
3. 交换有条件if (i < j) swap(q[i], q[j]);
4. 基值要固定x = q[l + r >> 1]
5. j最后的角标表示 l-j 是小于x的

二、归并排序

void merge_sort(int q[], int l, int r)
{
    if (l >= r) return;

    int mid = l + r >> 1;
    merge_sort(q, l, mid);
    merge_sort(q, mid + 1, r);

    int k = 0, i = l, j = mid + 1;
    while (i <= mid && j <= r)
        if (q[i] <= q[j]) tmp[k ++ ] = q[i ++ ];
        else tmp[k ++ ] = q[j ++ ];

    while (i <= mid) tmp[k ++ ] = q[i ++ ];
    while (j <= r) tmp[k ++ ] = q[j ++ ];

    for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];
}

1. 先分段
2. 用一个额外数组，汇总两个分数组的排序

1. 归并排序模板题 ✔ ✔1.31

★二刷总结

1. r = mid + 1
2. while if
3. temp 的k = 1
   a 的i = L

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();
        }
        merge_sort(0,n-1,nums);
        for (int i = 0; i < n; i++) {
            System.out.print(nums[i] + " ");
        }
    }
    
    public static void merge_sort (int l,int r,int[] nums) {
        if (l >= r) {
            return;
        }
        int mid = (l+r) >> 1;
        int i = l,j = mid + 1;
        merge_sort(l,mid,nums);
        merge_sort(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++];
            }
        }
        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++];
        }
    }
}

★2. 逆序对的数量 ✔ ✔1.31

原题链接

★二刷总结

1. java中new不能全new

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 ✔1.31

原题链接

★二刷总结(mid >= x 则是 输出最左边一个)

第一个大于等于x的数 || 最后一个大于等于x的数

mid < x 则是 往右

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 1e-8 ✔1.31

原题链接

二刷总结

1. 精确度是1e-8
2. l或者r直接等于mid

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 ✔1.31

原题链接
二刷：

1. 因为先计算小数，所以先把小数入 vector
2. 存放的时候先存放小数，这样i 一个角标就可以作为两个数的运算位置进行运算，如果相反的话，因为需要先计算最小值，那么就需要用两个角标指向最小值了
   

BigInteger

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 ✔1.31

原题链接

二刷：

1. 小减大 需要转换成 大-小
2. 如果出现负数 (x+10)%10 t = 1不是-1
3. t = a[i] - t; t = t - b[i]

a.subtract(b)

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：

1. 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 ✔(12分钟)2.1

原题链接

#include 
#include 
#include 

using namespace std;

vector<int> div(vector<int> &A, int b, int &r)
{
    vector<int> C;
    r = 0;
    for (int i = A.size() - 1; i >= 0; i -- )
    {
        r = r * 10 + A[i];
        C.push_back(r / b);
        r %= b;
    }
    reverse(C.begin(), C.end());
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

int main()
{
    string a;
    vector<int> A;

    int B;
    cin >> a >> B;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');

    int r;
    auto C = div(A, B, r);

    for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];

    cout << endl << r << endl;

    return 0;
}

五、前缀和S与差分a

1. 前缀和(2分钟)

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 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(7分钟)

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 m = scanner.nextInt();
        int q = scanner.nextInt();
        int[][] a = new int[n+1][m+1];
        int[][] s = new int[n+1][m+1];
        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= m; j++) {
                a[i][j] = scanner.nextInt();
            } 
        }
        for(int i = 1 ; i <= n  ; i ++ ){
            for(int j = 1 ;j <= m ; j ++ ){
                s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + a[i][j];
            }
        }
        while (q-->0) {
            int x1 = scanner.nextInt();
            int y1 = scanner.nextInt();
            int x2 = scanner.nextInt();
            int y2 = scanner.nextInt();
            System.out.println(s[x2][y2] - s[x1-1][y2] - s[x2][y1-1] + s[x1-1][y1-1]);
        }
    }
}

3. 差分(9分钟)

二刷总结

1. 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

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 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(12分钟)

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 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(20分钟)

二刷总结(以空间换时间)

原题链接

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(7分钟)

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 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(8分钟)

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 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

最全二进制算法总结

1. 位运算算法(2分钟)

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

sort(alls.begin(),alls.end());
alls.erase(unique(alls.begin(),alls.end()),alls.end());

去重 V.erase(unique(.begin(),.end()),.end());

1. ★ 区间和

原题链接

在草稿纸上列出需要几个数组，就清晰了

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(7分钟)

原题链接

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;
    }
    
}