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 }
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 rightwhile( 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 showedwhile( 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 showedwhile( left<=right ){
mid = left+(right-left)/2;if( a[mid]>k ){
right = mid-1;
last = mid;
}
else
left = mid+1;
}return last;
}
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;
}
分治法--二分查找、乘方、斐波那契数,布布扣,bubuko.com