折半查找又称为“二分查找”,算法复杂度为nlog2n。
1.折半查找思路
设有序顺序表{a[0], a[1], ......, a[n-1]},先求出查找区间中间元素下标mid,然后将该位置值a[mid]与要查找值key比较,比较结果有3种可能性:
- 若key=a[mid],则查找成功,返回下标;
- 若key<a[mid],则要查找元素在mid左侧,缩小查找区间到表前半部分再进行折半查找;
- 若key>a[mid],则要查找元素在mid右侧,缩小查找区间到表后半部分再进行折半查找。
2.过程示例
3.代码实现
①
非
递归
//非递归
public int binSearch(int a[], int low, int high, int key)
{
while(low<=high)
{
int mid=(low+high)/2;
if(a[mid]==key)
return mid;
else if(key<a[mid])
high=mid-1;
else
low=mid+1;
}
return -1;
}
②
非递归
//递归
public int binRecSearch(int a[], int low, int high, int key)
{
if(low<=high)
{
int mid=(low+high)/2;
if(a[mid]==key)
return mid;
else if(key<a[mid])
return binRecSearch(a, low,mid-1, key);
else
return binRecSearch(a,mid+1, high, key);
}
else
return -1;
}
4.完整代码示例
public class Test {
//非递归
public int binSearch(int a[], int low, int high, int key)
{
while(low<=high)
{
int mid=(low+high)/2;
if(a[mid]==key)
return mid;
else if(key<a[mid])
high=mid-1;
else
low=mid+1;
}
return -1;
}
//递归
public int binRecSearch(int a[], int low, int high, int key)
{
if(low<=high)
{
int mid=(low+high)/2;
if(a[mid]==key)
return mid;
else if(key<a[mid])
return binRecSearch(a, low,mid-1, key);
else
return binRecSearch(a,mid+1, high, key);
}
else
return -1;
}
public static void main(String[] args)
{
int[] a={ -36, -3, 5, 16, 24, 30, 78, 84, 345, 1004};
Test test=new Test();
//非递归
int pos1=test.binSearch(a,0,a.length-1,-3);
if(pos1!=-1)
System.out.println("-3所在的位置下标是 "+pos1);
else
System.out.println("找不到该数字84!");
//递归
int pos2=test.binRecSearch(a,0,a.length-1,84);
if(pos2!=-1)
System.out.println("84所在的位置下标是 "+pos2);
else
System.out.println("找不到该数字84!");
}
}
运行结果:
要记得一点,查找的序列首先必须是有序的才能进行折半查找。而
折半查找的递归和非递归思路都是差不多的。(凡星逝水2018)