手写常用算法

import java.util.Scanner;
/**
冒泡排序
*/
public class Solution{
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int[] a = {5,456,5,4,8,6,12,1};
        Bubble bubble = new Bubble();
        bubble.bubbleSort(a);
        for (int i = 0; i < a.length; i++) {
            System.out.print(a[i] + " ");
        }
    }
}

class Bubble {
    int[] bubbleSort(int[] a) {
        for (int i = 0; i < a.length - 1; i++) {
            for (int j = i; j < a.length; j++) {
                if(a[i] > a[j]) {
                    swap(a,i,j);
                }
            }
        }
        return a;
    }
    
    void swap(int[] a, int i, int j) {
        int tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
    }
}

class Solution {
    public void quickSort(int[] num, int left, int right) {
        int i, j, tmp;
        if(left > right) return;
        tmp = num[left]; //基准点
        i = left;
        j = right;
        while(i != j) {
            while(i<j && num[j] >= tmp) j--;
            while(i<j && num[i] <= tmp) i++;
            if(i < j) swap(num, i, j);
        }
        swap(num, left, j);
        quickSort(num, left, right - 1);
        quickSort(num, left + 1, right);
        return;
    }
    
    public static void swap(int[] num, int i, int j) {
        int tmp = num[i];
        num[i] = num[j];
        num[j] = tmp;
    }
    
    public static void main(String[] args) {
        Solution s = new Solution();
        int[] num = {2,1,5,3,4};
        s.quickSort(num, 0, 4);
        for(int i=0; i<5; i++) {
            System.out.println(num[i]);
        }
    }
}

1
2
3
4
5

class Solution{
    public static void main(String[] args) {
        int[] a = {15,4654,88,1,556,55,45};
        MergeSort mergeSort = new MergeSort();
        mergeSort.mergeSort(a, 0, a.length - 1);
        for (int i  = 0; i < a.length; i++) {
            System.out.print(a[i] + " ");
        }
    }
}

class MergeSort{
    
    public void merge(int[] a, int left, int mid, int right) {
        int[] tmp = new int[a.length];
        int p1 = left, p2 = mid + 1, k = left;
        while(p1 <= mid && p2 <= right) {
            if(a[p1] <= a[p2])
                tmp[k++] = a[p1++];
            else tmp[k++] = a[p2++];
        }
        while(p1<=mid) tmp[k++] = a[p1++];
        while(p2<=right) tmp[k++] = a[p2++];
        for(int i = left; i <= right; i++) {
            a[i] = tmp[i];
        }
    }
    public void mergeSort(int[] a, int st, int ed) {
        if(st < ed) {
            int mid = (st + ed) / 2;
            mergeSort(a, st, mid);
            mergeSort(a, mid+1, ed);
            merge(a, st, mid, ed);
        }
    }
}

import java.util.Scanner;
import java.util.Stack;

/**
 *十进制转换为二进制
*/
public class Solution{
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int num = cin.nextInt();
        Stack<Integer> s = new Stack<>();
        while(num != 0) {
            int tmp = num % 2;
            s.push(tmp);
            num >>= 1;
        }
        while(!s.isEmpty()) {
            System.out.print(s.peek());
            s.pop();
        }
    }
}

import java.util.Scanner;
/**
2的n次方的二进制数只有某一位是1，所以根据这个特点
当x&(x-1) == 0时则说明该数是2的n次方
 */
public class Solution{
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int num = cin.nextInt();
        boolean flag = (num & (num - 1)) == 0 && num != 0? true : false;
        System.out.println(flag);
    }
}

import java.util.Scanner;

public class Solution{
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int[] nums = {1,3,3,4,6,5};
        int n = nums.length, ans = 0;
        int[] hash = new int[n+1];
        for (int i = 0; i < n; i++) {
            if(hash[nums[i]] == 0) {
                hash[nums[i]] = 1;
            } else {
                ans = nums[i];
            }
        }
        System.out.println(ans);
    }
}

import java.util.Scanner;
/**
一个数组中，只有两个数字仅出现一次，其他数字均出现两次，找出这两个数字
*
* 要求：不借助任何空间 
* a ^ 0 = a,  a ^ a = 0，因此这里可以借助异或运算可以实现
*/
public class Solution{
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int[] nums = {1,2,1,6,8,6};
        int[] num1 = new int[5];
        int[] num2 = new int[5];
        int n = nums.length, ans = 0;
        int diff = 0;
        for (int i = 0; i < n; i++) {
            diff ^= nums[i];
        }
        diff &= -diff;
        for (int i : nums) {
            if((i & diff) == 0)
                num1[0] ^= i;
            else 
                num2[0] ^= i;
        }
        System.out.println(num1[0]);
        System.out.println(num2[0]);
    }
}