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分治法--二分查找、乘方、斐波那契数

2024-07-04 12:21:04   226人阅读

1、二分查找

常见错误：

  • 死循环：循环体外的初始化条件,与循环体内的迭代步骤, 都必须遵守一致的区间规则,也就是说,如果循环体初始化时,是以左闭右开区间为边界的,那么循环体内部的迭代也应该如此.如果两者不一致,会造成程序的错误.
  • 溢出：middle = left + (right - left) / 2
  • 终止条件：一般来说，如果左闭右闭，则left<=right； 如果一开一闭，则left<right； 关键看left能不能等于right，而且要考虑实际情况，有时不能这样简单终结，会出现死循环，如下面的binarySearch_better（）。
running time analysis：T(n) = T(n/2) +  Θ(1) ---> T(n) = Θ(lgn)
 
正确算法：
 1 //normal binarySearch without optimization, closed interval, return value is mid(maybe mid=left=right)
 2 int binarySearch(int a[], int n, int k){
 3     if(n<1) return -1;
 4 
 5     int left,right,mid;
 6     left = 0;       //closed left
 7     right = n-1;    //closed right
 8 
 9     while( left<=right ){
10         mid = left+(right-left)/2;
11 
12         if( a[mid]<k )  left = mid+1;
13         else if( a[mid]>k ) right = mid-1;
14         else return mid;
15     }
16     return -1;
17 }
View Code
减少比较次数为1次
//more efficient way, left closed right away interval, return value is mid=left
int binarySearch_better(int a[], int n, int k){
    if( n<1 )   return -1;

    int left,right, mid;
    left = 0;   //closed left
    right = n;  //open right

    while( left+1!=right ){
        mid = left+(right-left)/2;

        if( a[mid]>k )  right = mid;
        else left = mid;
    }

    if( a[left]!=k || left<0 )  return -1;
    return left;
}
 
//find the location that k first showed
int binarySearch_firstShow(int a[],int n, int k){
    if( n<1 ) return -1;

    int left,right,mid,last;
    left=0;     //closed left
    right=n-1;  //closed right
    last=-1;    //record the first place k showed

    while( left<=right ){
        mid = left+(right-left)/2;

        if( a[mid]>k )  right = mid-1;
        else if( a[mid]<k ) left = mid+1;
        else{
            right = mid-1;
            last = mid;
        }
    }

    return last;
}
 
//find the first location that a[location]>k
int binarySearch_firstBig(int a[],int n, int k){
    if( n<1 ) return -1;

    int left,right,mid,last;
    left=0;     //closed left
    right=n-1;  //closed right
    last=-1;    //record the first place k showed

    while( left<=right ){
        mid = left+(right-left)/2;

        if( a[mid]>k ){
            right = mid-1;
            last = mid;
        }
        else
            left = mid+1;
    }

    return last;
}
View Code

2、斐波那契数

running time analysis：
  • 采用递归，f(n) = f(n-1)+f(n-2)，没有减少问题规模，反而扩大了，因为f(n-1)和f(n-2)的计算有重复的部分。
  • 从f1一个一个加过来计算，时间为Θ(n)；
  • 缩减为非线性时间，采用规律：{f(n+1),f(n),f(n),f(n-1)} <==> a{1,1,1,0}^n，利用分治法，T(n) = T(n/2) + Θ(1) = Θ(lgn)；
//{f(n+1),f(n),f(n),f(n-1)} <==> a{1,1,1,0}^n, return value is f(n)
long long fibonacci( long long a[2][2],int n){
    if(n==1){
        a[0][0] = a[0][1] = a[1][0] = 1;
        a[1][1] = 0;
        return a[0][1];
    }
    long long b[2][2];
    fibonacci(b,n/2);
    for( int i=0; i<2; i++ )
    for( int j=0; j<2; j++ ){
        a[i][j] = 0;
        for( int k=0; k<2; k++ )
            a[i][j] += b[i][k]*b[k][j];
    }
    if( n%2==0 )    return a[0][1];
    //n is odd, then a = a*{1,1,1,0};
    int t1,t2;
    t1 = a[0][0];
    t2 = a[0][1];
    a[0][0] = t1+t2;
    a[0][1] = a[1][0] = t1;
    a[1][1] = t2;
    return a[0][1];
}

 

3、乘方

running time analysis: T(n) = T(n/2) + Θ(1) = Θ(lgn)
//compute x^n.
long long power(int x, int n){
    if( n==1 )  return (long long)x;

    long long half = power(x,n/2);
    if( n/2*2 == n )   return half*half;
    return half*half*x;
}

 

  


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