基本数据操作

1.数组逆序

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        int arr[] = {0,57,7,5,1};
        int N = arr.length;
        for(int i=0;i<N/2;i++){
            int tmp = arr[i];
            arr[i] = arr[N-i-1];
            arr[N-i-1] = tmp;
        }
        System.out.println(Arrays.toString(arr));
    }
}

2.判断是否是素数

public static boolean isPrime(int a) {
        if (a < 2){
            return false;
        }
        else {
            for (int i = 2; i <= Math.sqrt(a); i++){
                if (a % i == 0) return false;
            }
        }
        return true;
    }

3.开根号

/*
     * 牛顿迭代法        开方     2 = 根号2
     */
    public static double sqrt(double num){
        if(num<0){
            return Double.NaN;
        }
        double err = 1e-15;
        double t = num;
        while(Math.abs(t-num/t)> err*t){
            t = (t+num/t)/2.0;
            
        }
        return t;
    }

4. 二分查找,注意查询之前要排序

public static void main(String[] args) {
        int arr[] = {0,57,7,5,1};
        Arrays.sort(arr);
        System.out.println(Arrays.toString(arr));
        System.out.println(rank(7,arr));
    }
    
    public static int rank(int key,int[] arr){
        return rank(key,arr, 0, arr.length-1);
    }
    
    public static int rank(int key,int [] arr, int lo,int hi){
        //return 0;
        if(lo>hi){
            return -1;
        }
        int mid = lo+(hi-lo)/2;
        if(arr[mid]>key){
            return rank(key,arr,lo,mid-1);
        }else if(arr[mid]<key){
            return rank(key,arr,mid+1,hi);
        }else{
            return mid;
        }
        //return hi;
    }

 4.判断  回文数

/**
     * 判断是不是回文数
     * @param str
     * @return
     */
    public static boolean isBack(String str){
        int N = str.length();
        for(int i=0;i<N/2;i++){
            if(str.charAt(i)!=str.charAt(N-i-1)){
                return false;
            }
        }
        return true;     
    }

 

posted on 2017-09-13 16:56  biyangqiang  阅读(119)  评论(0编辑  收藏  举报

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