问题
I have been using my time off university to practice Java through coding algorithms. One of the algorithms I coded was the binary search:
public class BinarySearch {
private static int list[] = {3, 6, 7, 8, 9, 10};
public static void main(String[] args) {
BinarySearch b = new BinarySearch();
b.binarySearch(list);
}
public void binarySearch(int[] args) {
System.out.println("Binary search.");
int upperBound = args.length;
int lowerBound = 1;
int midpoint = (upperBound + lowerBound) / 2;
int difference = upperBound - lowerBound;
int search = 7;
for (int i = 0; i < args.length; i++) {
if (search < args[midpoint - 1] && difference != 1) {
upperBound = midpoint - 1;
midpoint = upperBound / 2;
} else if (search > args[midpoint - 1] && difference != 1) {
lowerBound = midpoint + 1;
midpoint = (lowerBound + upperBound) / 2;
} else if (search == args[midpoint - 1]) {
midpoint = midpoint - 1;
System.out.println("We found " + search + " at position " + midpoint + " in the list.");
i = args.length;
} else {
System.out.println("We couldn't find " + search + " in the list.");
i = args.length;
}
}
}
}
I really want to be able to write a much cleaner and efficient binary search algorithm, an alternative to what I've coded. I have seen examples of how recursion is used such as when doing factorial with numbers which I understand. However when coding something of this complexity I am confused on how to use it to my advantage. Therefore my question is how do I apply recursion when coding a binary search algorithm. And if you have any tips for me to perfect my recursion skills even if it has to be something that doesn't regard to binary search then please feel free to post.
回答1:
If you really want to use recursion, this should do it.
public static int binarySearch(int[] a, int target) {
return binarySearch(a, 0, a.length-1, target);
}
public static int binarySearch(int[] a, int start, int end, int target) {
int middle = (start + end) / 2;
if(end < start) {
return -1;
}
if(target==a[middle]) {
return middle;
} else if(target<a[middle]) {
return binarySearch(a, start, middle - 1, target);
} else {
return binarySearch(a, middle + 1, end, target);
}
}
回答2:
Here is an easier way of doing binary search:
public static int binarySearch(int intToSearch, int[] sortedArray) {
int lower = 0;
int upper = sortedArray.length - 1;
while (lower <= upper) {
int mid = lower + (upper - lower) / 2;
if(intToSearch < sortedArray[mid])
upper = mid - 1;
else if (intToSearch > sortedArray[mid])
lower = mid + 1;
else
return mid;
}
return -1; // Returns -1 if no match is found
}
回答3:
Following is a code sample extracted from here.
public class BinarySearch {
public boolean find(int[] sortedValues, int value) {
return search(sortedValues, value, 0, sortedValues.length - 1);
}
private boolean search(int[] sorted, int value, int leftIndex, int rightIndex) {
// 1. index check
if (leftIndex > rightIndex) {
return false;
}
// 2. middle index
int middle = (rightIndex + leftIndex) / 2;
// 3. recursive invoke
if (sorted[middle] > value) {
return search(sorted, value, leftIndex, middle - 1);
} else if (sorted[middle] < value) {
return search(sorted, value, middle + 1, rightIndex);
} else {
return true;
}
}
}
You can find implementations of the below test cases against the above binary search implementation as well in the reference link.
1. shouldReturnFalseIfArrayIsEmpty()
2. shouldReturnFalseIfNotFoundInSortedOddArray()
3. shouldReturnFalseIfNotFoundInSortedEvenArray()
4. shouldReturnTrueIfFoundAsFirstInSortedArray()
5. shouldReturnTrueIfFoundAtEndInSortedArray()
6. shouldReturnTrueIfFoundInMiddleInSortedArray()
7. shouldReturnTrueIfFoundAnywhereInSortedArray()
8. shouldReturnFalseIfNotFoundInSortedArray()
回答4:
A possible example is :
// need extra "helper" method, feed in params
public int binarySearch(int[] a, int x) {
return binarySearch(a, x, 0, a.length - 1);
}
// need extra low and high parameters
private int binarySearch(int[ ] a, int x,
int low, int high) {
if (low > high) return -1;
int mid = (low + high)/2;
if (a[mid] == x) return mid;
else if (a[mid] < x)
return binarySearch(a, x, mid+1, high);
else // last possibility: a[mid] > x
return binarySearch(a, x, low, mid-1);
}
Here you can check in C Binary Search, With and Without Recursion
Source : http://www.cs.utsa.edu/~wagner/CS3343/recursion/binsearch.html
回答5:
Here is a algorithm which should get you going. Let your method signature be:
public boolean binarysearchRecursion(Array, begin_index,end_index, search_element)
- Check if your begin_index > end_index if YES then return
false. - Calculate
mid_elementfor your input array. - Check if your
search_elementis equal to thismid_element. if YES returntrue - If
mid_element>search_elementCall your method with forrange 0 - mid - If
mid_element<search_elementCall your method with for rangemid+1 - Length_of_Array
Also as @DwB said in his comment you are better using loop to get things done. Some problems are recursive in nature(Like binary tree problems). But this one is not one of them.
回答6:
This is another way of doing recursion:
int[] n = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
@Test
public void testRecursiveSolution() {
Assert.assertEquals(0, recursiveBinarySearch(1,n));
Assert.assertEquals(15, recursiveBinarySearch(16,n));
Assert.assertEquals(14, recursiveBinarySearch(15,n));
Assert.assertEquals(13, recursiveBinarySearch(14,n));
Assert.assertEquals(12, recursiveBinarySearch(13,n));
Assert.assertEquals(11, recursiveBinarySearch(12,n));
Assert.assertEquals(10, recursiveBinarySearch(11,n));
Assert.assertEquals(9, recursiveBinarySearch(10,n));
Assert.assertEquals(-1, recursiveBinarySearch(100,n));
}
private int recursiveBinarySearch(int n, int[] array) {
if(array.length==1) {
if(array[0]==n) {
return 0;
} else {
return -1;
}
} else {
int mid = (array.length-1)/2;
if(array[mid]==n) {
return mid;
} else if(array[mid]>n) {
return recursiveBinarySearch(n, Arrays.copyOfRange(array, 0, mid));
} else {
int returnIndex = recursiveBinarySearch(n, Arrays.copyOfRange(array, mid+1, array.length));
if(returnIndex>=0) {
return returnIndex+mid+1;
} else {
return returnIndex;
}
}
}
}
回答7:
While it doesn't return the index, this at least returns the idea of 'yes' or 'no' that something is in the collection:
public static boolean recursive(int[] input, int valueToFind) {
if (input.length == 0) {
return false;
}
int mid = input.length / 2;
if (input[mid] == valueToFind) {
return true;
} else if (input[mid] > valueToFind) {
int[] smallerInput = Arrays.copyOfRange(input, 0, mid);
return recursive(smallerInput, valueToFind);
} else if (input[mid] < valueToFind) {
int[] smallerInput = Arrays.copyOfRange(input, mid+1, input.length);
return recursive(smallerInput, valueToFind);
}
return false;
}
回答8:
A recursion BinarySearch with break conditions in case you can not find the value you are looking for
public interface Searcher{
public int search(int [] data, int target, int low, int high);
}
The Implementation
public class BinarySearch implements Searcher {
public int search(int[] data, int target, int low, int high) {
//The return variable
int retorno = -1;
if(low > high) return retorno;
int middle = (high + low)/2;
if(target == data[middle]){
retorno = data[middle];
}else if(target < data[middle] && (middle - 1 != high)){
//the (middle - 1 != high) avoids beeing locked inside a never ending recursion loop
retorno = search(data, target, low, middle - 1);
}else if(target > data[middle] && (middle - 1 != low)){
//the (middle - 1 != low) avoids beeing locked inside a never ending recursion loop
retorno = search(data, target, middle - 1, high);
}else if(middle - 1 == low || middle - 1 == high){
//Break condition if you can not find the desired balue
retorno = -1;
}
return retorno;
}
}
来源：https://stackoverflow.com/questions/19012677/how-to-use-recursion-in-creating-a-binary-search-algorithm