无处不在的二分查找



问题1描述

给一个已经排序的数组,其中有N个互不相同的元素。要求使用最小的比较次数找出其中的一个元素。(你认为二分查找在排序数组里找一个元素是最优的算法的吗?)
不需要太多的理论,这是一个典型的二分查找算法。先看下面的代码:
int BinarySearch(int A[], int l, int r, int key)
03{
04    int m;
05    while( l <= r )
06    {
07        m = l + (r-l)/2;
08 
09        if( A[m] == key ) //第一次比较
10            return m;
11 
12        if( A[m] < key ) // 第二次比较
13            l = m + 1;
14        else
15            r = m - 1;
16    }
17 
18    return -1;
19}
理论上,我们最多需要 logN+1 次比较。仔细观察,我们在每次迭代中使用两次比较,除了最后比较成功的一次。实际应用上,比较也是代价高昂的操作,往往不是简单的数据类型的比较。减少比较的次数也是优化的方向之一。
下面是一个比较次数更少的实现:
在while循环中,我们仅依赖于一次比较。搜索空间( l->r )不断缩小,我们需要一个比较跟踪搜索状态。
需要注意的,要保证我们恒等式(A[l] <= key & A[r] > key)正确,后面还会用到循环不变式。
04int BinarySearch(int A[], int l, int r, int key)
05{
06    int m;
07    while( r - l > 1 )
08    {
09        m = l + (r-l)/2;
10 
11        if( A[m] <= key )
12            l = m;
13        else
14            r = m;
15    }
16    if( A[l] == key )
17        return l;
18    else
19        return -1;
20}



问题4描述

有一个已排序的数组(无相同元素)在未知的位置进行了旋转操作,找出在新数组中的最小元素所在的位置。
例如:原数组 {1,2,3,4,5,6,7,8,9,10}, 旋转后的数组可能是 {6,7,8,9,10, 1,2,3,4,5 },也可能是 {8,9,10,1,2,3,4,5,6,7 }
我们不断的缩小 l 左指针和 r 右指针直到有一个元素。把上面划横线的作为第一部分,剩下的为第二部分。如果中间位置m落在第一部分,即A[m] < A[r] 不成立,我们排序掉区间 A[m+1 ... r]。 如果中间位置m落在第二部分,即 A[m]<A[r]成立,我们缩小区间至 A[m+1 .... r ]。 直到搜索的区间大小为1就结束。
int BinarySearchIndexOfMinimumRotatedArray(int A[], int l, int r)
02{
03    int m;
04 
05    // 先决条件: A[l] > A[r]
06    if( A[l] <= A[r] )
07        return l;
08 
09    while( l <= r )
10    {
11        //终止条件
12        if( l == r )
13            return l;
14 
15        m = l + (r-l)/2; // 'm' 可以落在第一部分或第二部分
16 
17        if( A[m] < A[r] )
18            // (m < i <= r),可以排除 A[m+1 ... r]
19            r = m;
20        else
21            // min肯定在区间 (m < i <= r),
22            // 缩小区间至 A[m+1 ... r]
23            l = m+1;
24    }
25    return -1;
26}
27 
28int BinarySearchIndexOfMinimumRotatedArray(int A[], int size)
29{
30    return BinarySearchIndexOfMinimumRotatedArray(A, 0, size-1);
31}




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