1.概述。

In this article, we’ll cover advantages of a binary search over a simple linear search and walk through its implementation in Java.

2.需要高效的搜索。

Let’s say we’re in the wine-selling business and millions of buyers are visiting our application every day.

Through our app, a customer can filter out items which have a price below n dollars, select a bottle from the search results, and add them to his cart. We have millions of users seeking wines with a price limit every second. The results need to be fast.

On the backend, our algorithm runs a linear search through the entire list of wines comparing the price limit entered by the customer with the price of every wine bottle in the list.

Then, it returns items which have a price less than or equal to the price limit. This linear search has a time complexity of O(n).

This means the bigger the number of wine bottles in our system, the more time it will take. The search time increases proportionately to the number of new items introduced.

If we start saving items in sorted order and search for items using the binary search, we can achieve a complexity of O(log n).

With binary search, the time taken by the search results naturally increases with the size of the dataset, but not proportionately.

3.二进制搜索。

Simply put, the algorithm compares the key value with the middle element of the array; if they are unequal, the half in which the key cannot be part of is eliminated, and the search continues for the remaining half until it succeeds.

Remember – the key aspect here is that the array is already sorted.

If the search ends with the remaining half being empty, the key is not in the array.

3.1.迭代实施。

``````public int runBinarySearchIteratively(
int[] sortedArray, int key, int low, int high) {
int index = Integer.MAX_VALUE;

while (low <= high) {
int mid = low  + ((high - low) / 2);
if (sortedArray[mid] < key) {
low = mid + 1;
} else if (sortedArray[mid] > key) {
high = mid - 1;
} else if (sortedArray[mid] == key) {
index = mid;
break;
}
}
return index;
}``````

The runBinarySearchIteratively method takes a sortedArray, key & the low & high indexes of the sortedArray as arguments. When the method runs for the first time the low, the first index of the sortedArray, is 0, while the high, the last index of the sortedArray, is equal to its length – 1.

runBinarySearchIteratively方法以sortedArraykeysortedArraylowhigh索引为参数。当该方法第一次运行时，low，即sortedArray的第一个索引，是0，而high，即sortedArray的最后一个索引，等于其长度-1。

The middle is the middle index of the sortedArray. Now the algorithm runs a while loop comparing the key with the array value of the middle index of the sortedArray.

middlesortedArray的中间索引。现在，该算法运行一个while循环，将keysortedArray的中间索引的数组值进行比较。

Notice how the middle index is generated (int mid = low + ((high – low) / 2). This to accommodate for extremely large arrays. If the middle index is generated simply by getting the middle index (int mid = (low + high) / 2), an overflow may occur for an array containing 230 or more elements as the sum of low + high could easily exceed the maximum positive int value.

3.2.递归实现。

Now, let’s have a look at a simple, recursive implementation as well:

``````public int runBinarySearchRecursively(
int[] sortedArray, int key, int low, int high) {
int middle = low  + ((high - low) / 2);

if (high < low) {
return -1;
}

if (key == sortedArray[middle]) {
return middle;
} else if (key < sortedArray[middle]) {
return runBinarySearchRecursively(
sortedArray, key, low, middle - 1);
} else {
return runBinarySearchRecursively(
sortedArray, key, middle + 1, high);
}
}
``````

The runBinarySearchRecursively method accepts a sortedArray, key, the low and high indexes of the sortedArray.

runBinarySearchRecursively方法接受一个sortedArraykey、lowhigh的索引。

3.3.使用Arrays.binarySearch() 。

``````int index = Arrays.binarySearch(sortedArray, key);
``````

A sortedArray and an int key, which is to be searched in the array of integers, are passed as arguments to the binarySearch method of the Java Arrays class.

A sortedArray和一个int key（要在整数数组中搜索）被作为参数传递给Java Arrays类的binarySearch方法。

3.4.使用Collections.binarySearch()。

``````int index = Collections.binarySearch(sortedList, key);
``````

A sortedList & an Integer key, which is to be searched in the list of Integer objects, are passed as arguments to the binarySearch method of the Java Collections class.

A sortedList & 一个Integer key，要在Integer对象的列表中进行搜索，作为参数传递给Java Collections类的binarySearch方法。

3.5.性能。

Whether to use a recursive or an iterative approach for writing the algorithm is mostly a matter of personal preference. But still here are a few points we should be aware of:

1. Recursion can be slower due to the overhead of maintaining a stack and usually takes up more memory
2. Recursion is not stack-friendly. It may cause StackOverflowException when processing big data sets
3. Recursion adds clarity to the code as it makes it shorter in comparison to the iterative approach

1.由于维护堆栈的开销，递归可能会更慢，通常会占用更多的内存

2.递归对堆栈不友好。在处理大数据集时，它可能导致StackOverflowException
3.
3.递归增加了代码的清晰度，因为与迭代方法相比，它使代码更短。

Ideally, a binary search will perform less number of comparisons in contrast to a linear search for large values of n. For smaller values of n, the linear search could perform better than a binary search.

One should know that this analysis is theoretical and might vary depending on the context.

Also, the binary search algorithm needs a sorted data set which has its costs too. If we use a merge sort algorithm for sorting the data, an additional complexity of n log n is added to our code.

So first we need to analyze our requirements well and then take a decision on which search algorithm would suit our requirements best.

4.结论。

This tutorial demonstrated a binary search algorithm implementation and a scenario where it would be preferable to use it instead of a linear search.

Please find the code for the tutorial over on GitHub.