Java implementation of binary search algorithm

Algorithm

If there is a set of numbers for 3,12,24,36,55,68,75,88 to look up a given value of 24. Three variables can be set front,mid,end respectively to the upper bound of the data, the middle and lower bounds, mid= (front+end)/2.

1. Start front=0 (point 3), end=7 (Point 88), then mid=3 (point to 36). Because of mid>x, it should be found in the first half of the paragraph. 2. Make the new end=mid-1=2, while the front=0 is unchanged, then the new mid=1. At this point x>mid, it should be found in the second half of the paragraph. 3. Make the new front=mid+1=2, and end=2 unchanged, then the new mid=2, at this time a[mid]=x, find success. If the number to be looked up is not the number in the series, for example X=25, when the third judgment, X>a[mid], according to the above rules, so that front=mid+1, that is, front=3, the case of front>end, indicating that the search is unsuccessful.

Example: Find the data x that the user has entered into an array with n elements in order. the algorithm is as follows:

1. Determine the lookup range front=0,end=n-1, calculate the item mid= (front+end)/2.

2. If a[mid]=x or front>=end, end the lookup; otherwise, continue down.

3. If the a[mid]<x indicates that the element value to be found is only possible in the range larger than the middle item element, assign the value of Mid+1 to front and recalculate mid, go to step 2, and if a[mid]>x, the element value to be found may only be within the range smaller than the middle item element, Assign the value of Mid-1 to end and recalculate mid to go to step 2.

[one-dimensional array, binary find]2 algorithm complexity analysis time complexity

1. Worst case finding last element (or first element) Master theorem t (n) =t (N/2) +o (1) so T (n) =o (LOGN)
2. Best case Find Intermediate element O (1) The element that is found is the middle element (the middle of the odd-length sequence, the left-hand element of the even-length sequence)

Complexity of space:

S (n) =n

Java Implementation Code

Package Com.leo.kang.interview;public class BinarySearch {//Lookup count static int count;/** * @param args */public static void M Ain (string[] args) {//TODO auto-generated method stubint[] Array = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; System.out.println (searchrecursive (array, 0, array.length-1, 9)); System.out.println (count); count = 0; System.out.println (Searchloop (array, 9)); System.out.println (count);}            /** * Performs a recursive binary lookup that returns the position of the first occurrence of the value * * @param array * Sorted array * @param start * Start position * @param end * End position * @param findvalue * The value to be found * @return The position of the value in the array, starting with 0. Cannot find return-1 */public static int searchrecursive (int[] array, int start, int end,int findvalue) {///If the array is empty, return 1 directly, that is, find failed if (ARRA y = = null) {return-1;} Count++;if (start <= end) {//middle position int middle = (start + end)/1;//median int middlevalue = array[middle];if (Findvalue = = Middlevalue) {//Equals middle value returns return middle directly,} else if (Findvalue < Middlevalue) {///less than median value in front of median value find return searchrecursive (array , start, middle-1, findValue);} else {//greater than median after median value find return searchrecursive (array, middle + 1, end, findvalue);}} else {//return-1, that is, find failed return-1;}} /** * Loop binary lookup, returns the position of the first occurrence of the value * * @param array * Sorted array * @param findvalue * Required value * @return value in the position of the array, from 0 start. Cannot find return-1 */public static int searchloop (int[] array, int findvalue) {//If the array is empty, return 1 directly, that is, find failed if (array = = null) {return-1;} Start position int start = 0;//end position int end = Array.length-1;while (start <= end) {count++;//middle position int middle = (start + end)  /2;//middle value int middlevalue = array[middle];if (findvalue = = Middlevalue) {//equals median returns return middle;} else if (Findvalue < Middlevalue) {///less than median value is found in front of the median value, end = Middle-1;} else {//greater than median is found after the median value start = middle + 1;}} Returns-1, which is the lookup failed return-1;}}

　　

Java implementation of binary search algorithm