二分查找
折半查找过程可用二叉树来描述，把有序表中间位置上的结点作为树的根结点，左子表和右子表分别对应树的左子树和右子树。折半查找的过程就是走一条从根节点到被查结点的一条路径，比较的次数就是该路径中结点的个数，即，该结点在树中的层数。 深度为K的二叉树最多有2的K次方-1个结点，深度为7的二叉树最多有127个结点.
代码
public class BinarySearch {
public static int noRecursion(int[] arr, int key, int low, int high) {
while (low < high) {
int mid = (low + high) >>> 1;
int sortNum = arr[mid];
if (key > sortNum) {
low = mid + 1;
} else if (key < sortNum) {
high = mid - 1;
} else {
return mid;
}
}
return -1;
}
public static int recursion(int[] arr, int key, int low, int high) {
int mid = (low + high) >>> 1;
int sortNum = arr[mid];
int res = -1;
if (key > sortNum) {
res = recursion(arr, key, mid + 1, high);
} else if (key < sortNum) {
res = recursion(arr, key, low, mid - 1);
} else {
return mid;
}
return res;
}
}
复杂度
image.png
整个有序表可以生成深度为多少的二叉树
int TestBinSearch(int first, int end)
{
if(first < end)
{
int mid = (first + end) / 2;
int numOfLeftCmp = TestBinSearch(first, mid - 1);
int numOfRightCmp = TestBinSearch(mid + 1, end);
if(numOfLeftCmp > numOfRightCmp)
return numOfLeftCmp + 1;
else
return numOfRightCmp + 1;
}
else if(first == end)
return 1;
else
return 0;
}
