java算法每日一练2021/1/23

1.顺序查找

/**
 * 顺序查找
 * 时间复杂度O(n)
 * @param args
 */
public static void main(String[] args) {
    int[] nums = {1,1,5,13,6,9,8};
    System.out.println("该数值所在下标为："+search(nums,13));
}
public static int search(int[] nums, int num){
    int length = nums.length;
    for(int i = 0; i < length; i++){
        if(num == nums[i]){
            return i;
        }
    }
    return -1;
}
2.二分查找

/**
 * 二分查找 时间复杂度O(logn)，存储结构必须是顺序存储 ，数据有序
 * @param args
 */
public static void main(String[] args) {
    int[] nums = {1,2,3,6,7,8,10};
    System.out.println("该数值所在下标为："+search(nums,13));
}
public static int search(int[] nums, int num){
    int low = 1;
    int high = nums.length-1;
    while(low<=high) {
        int mid = (low+high)/2;
        if(nums[mid]<num)
            low=mid+1;   //要+1
        else if(nums[mid]>num)
            high=mid-1;  //要-1
        else
            return mid;
    }
    return -1;
}
3.插值查找

/**
 * 插值查找，对于表长较大，关键字分布比较均匀的查找表来说，可以采用
 * @param args
 */
public static void main(String[] args) {
    int[] nums = {1,2,3,6,7,8,10};
    System.out.println("该数值所在下标为："+search(nums,7));
}
public static int search(int[] nums, int num){
    int low = 1;
    int high = nums.length-1;
    while(low<=high) {
        int mid = low + (high - low) * (num - nums[low]) / (nums[high] - nums[low]);/*插值*/
        if(nums[mid]<num)
            low=mid+1;   //要+1
        else if(nums[mid]>num)
            high=mid-1;  //要-1
        else
            return mid;
    }
    return -1;
}
4.斐波那契查找

/*
 * 斐波那契数列
 * 采用递归
 */
public static int fib(int n) {
    if(n==0)
        return 0;
    if(n==1)
        return 1;
    return fib(n-1)+fib(n-2);
}

/*
 * 斐波那契数列
 * 不采用递归
 */
public static int fib2(int n) {
    int a=0;
    int b=1;
    if(n==0)
        return a;
    if(n==1)
        return b;
    int c=0;
    for(int i=2;i<=n;i++) {
        c=a+b;
        a=b;
        b=c;
    }
    return c;
}

/*
 * 斐波那契查找
 */
public static int fibSearch(int[] arr,int n,int key) {
    int low=1;  //记录从1开始
    int high=n;     //high不用等于fib(k)-1，效果相同
    int mid;

    int k=0;
    while(n>fib(k)-1)    //获取k值
        k++;
    int[] temp = new int[fib(k)];   //因为无法直接对原数组arr[]增加长度，所以定义一个新的数组
    System.arraycopy(arr, 0, temp, 0, arr.length); //采用System.arraycopy()进行数组间的赋值
    for(int i=n+1;i<=fib(k)-1;i++)    //对数组中新增的位置进行赋值
        temp[i]=temp[n];

    while(low<=high) {
        mid=low+fib(k-1)-1;
        if(temp[mid]>key) {
            high=mid-1;
            k=k-1;  
        }else if(temp[mid]<key) {
            low=mid+1;
            k=k-2;  
        }else {
            if(mid<=n)
                return mid;
            else
                return n;       //当mid位于新增的数组中时，返回n
        }
    }
    return 0;
}
public static void main(String[] args) {
    int[] arr = {0,1,16,24,35,47,59,62,73,88,99};
    int n=10;
    int key=59;
    System.out.println(fibSearch(arr, n, key)); 
}