zoukankan      html  css  js  c++  java
  • Java经典递归算法

    斐波那契数列

    package com.luna.base;
    public class BirthRabbit {
        public static void main(String[] args) {
            int i = 1;
            for (i = 1; i <= 20; i++) {
                System.out.println("兔子第" + i + "个月的总数为:" + f(i));
            }
        }
        public static int f(int x) {
            if (x == 1 || x == 2) {
                return 1;
            } else {
                return f(x - 1) + f(x - 2);
            }
        }
    }

    从1到100相加

    package com.luna.base;
    public class Plus {
        public int sum(int i) {
            if (i == 1) {
                return 1;
            }
            return i + sum(i - 1);
        }
        public static void main(String[] args) {
            Plus plus = new Plus();
            System.out.println("计算结果:" + plus.sum(100) + "!");
        }
    }

    100的阶乘

    package com.luna.base;
    import java.math.BigInteger;
    public class LoopMutiply {
        public BigInteger sum(int i) {
            if (i == 1) {
                return BigInteger.ONE;
            }
            return BigInteger.valueOf(i).multiply(sum(i - 1));
        }
        public static void main(String[] args) {
            LoopMutiply test = new LoopMutiply();
            try {
                System.out.println("计算结果:" + test.sum(50) + "!");
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
    }

    有序数组a、b合并成一个新的有序数组

    package com.luna.base;
    public class ArraySort {
        public static void main(String[] args) {
            int[] a = { 1, 3, 4 };
            int[] b = { 2, 3, 5, 6 };
            int[] c = mergeArray(a, b);
            for (int n : c) {
                System.out.print(n + " ");
            }
        }
        // 合并数组
        public static int[] mergeArray(int[] a, int[] b) {
            int result[] = new int[a.length + b.length];
            if (checkSort(a) && checkSort(b)) {
                // 说明ab数组都是有序的数组
                // 定义两个游标
                int i = 0, j = 0, k = 0;
                while (i < a.length && j < b.length) {
                    if (a[i] <= b[j]) {
                        result[k++] = a[i++];
                    } else {
                        result[k++] = b[j++];
                    }
                }
                while (i < a.length) {
                    // 说明a数组还有剩余
                    result[k++] = a[i++];
                }
                while (j < b.length) {
                    result[k++] = b[j++];
                }
            }
            return result;
        }
        // 检查一个数组是否是有序1 2 3
        public static boolean checkSort(int[] a) {
            boolean flag = false;// 默认不是有序的
            for (int i = 0; i < a.length - 1; i++) {
                if (a[i] > a[i + 1]) {
                    // 说明不是有序的
                    flag = false;
                    break;
                } else {
                    flag = true;
                }
            }
            return flag;
        }
    }

    归并排序算法实现

    package com.luna.base;
    public class MergingSort {    
        public static void sort(int[] data, int left, int right) {    
            if (left < right) {    
                // 首先找出中间的索引    
                int center = (left + right) / 2;    
        
                // 对中间索引左边的数组进行递归    
                sort(data, left, center);    
        
                // 对中间索引右边的数组进行递归    
                sort(data, center + 1, right);    
                // 合并    
                merge(data, left, center, right);    
            }    
        }    
        
        public static void merge(int[] data, int left, int center, int right) {    
            int[] tmpArr = new int[data.length];        
            int mid = center + 1;      
            // third记录中间数组的索引    
            int third = left;    
            int tmp = left;        
            while (left <= center && mid <= right) {    
                // 将两个数组中取出最小的数放入中间数组    
                if (data[left] <= data[mid]) {    
                    tmpArr[third++] = data[left++];    
                } else {    
                    tmpArr[third++] = data[mid++];    
                }    
            }    
        
            // 剩余部分依次放入中间数组    
            while (mid <= right) {    
                tmpArr[third++] = data[mid++];    
            }       
            while (left <= center) {    
                tmpArr[third++] = data[left++];    
            }                
            while(tmp <= right){    
                data[tmp] = tmpArr[tmp++];    
            }    
        }    
        
        public static void main(String[] args) {    
            int[] a = { 3, 2, 5, 4 };               
            sort(a, 0, a.length - 1);        
            for (int i = 0; i < a.length; i++) {    
                System.out.print(a[i] + " ");    
            }    
        }    
    }
    

    九九

    public static void main(String[] args) {
    for (int i = 1 ; i <= 9 ; i++){
    for (int j = 1 ; j <= i ; j++){
    System.out.print(i + "*" + j + "=" + i * j + " ");
    }
    System.out.println();
    }
    }
  • 相关阅读:
    win 10打开administrator
    Navicat
    mkpasswd
    恢复不小心删除的文件
    mysql双主出现1602错误
    scp远程拷贝文件免密办法
    iredmail邮箱服务器部署
    keepalived+nginx后端服务器access_log出现127.0.0.1的访问记录
    lsof 简介
    Codeforces #536 div2 E (1106E)Lunar New Year and Red Envelopes (DP)
  • 原文地址:https://www.cnblogs.com/gaobing1252/p/11110935.html
Copyright © 2011-2022 走看看