Binary Search

原创转载请注明出处：http://agilestyle.iteye.com/blog/2273610

 

Binary Search原理

package org.fool.test;

public class BinarySearch {

	public static int binarySearch(int[] nums, int num) {
		int low = 0;
		int high = nums.length - 1;

		while (low <= high) {
			int mid = (low + high) / 2;

			if (num > nums[mid]) {
				low = mid + 1;
			} else if (num < nums[mid]) {
				high = mid - 1;
			} else {
				return mid;
			}
		}

		return -1;
	}

	public static void main(String[] args) {

		int[] nums = { 12, 23, 34, 45, 56, 67, 77, 89, 90 };

		System.out.println(binarySearch(nums, 12));
		System.out.println(binarySearch(nums, 45));
		System.out.println(binarySearch(nums, 67));
		System.out.println(binarySearch(nums, 89));
		System.out.println(binarySearch(nums, 99));
	}
}

 

get square root of a number

核心思想：二分查找

1.define an initial range of (start, end)

2.keep testing mid^2 to approach value

3.update start or end to mid

4.stop when certain precision is achieved or exact square root is computed

package org.fool.java.test;

public class SqrtTest {
    public static void main(String[] args) {
        System.out.println("sqrt of 50 is " + sqrt(50));
        System.out.println("sqrt of 81 is " + sqrt(81));
        System.out.println("sqrt of 99 is " + sqrt(99));
    }

    private static double sqrt(double a) {
        if (a < 0) {
            return -1;
        }

        if (a == 0 || a == 1) {
            return a;
        }

        double precision = 0.00000001;
        double start = 0;
        double end = a;

        // define these 2 start/end values because usually 0 < sqrt(a) < a
        // however, if a < 1, then 0 < a < sqrt(a)
        if (a < 1) {
            end = 1;
        }

        // define a loop to continue if the precision is not yet achieved
        while (end - start > precision) {
            double mid = (start + end) / 2;
            double midSqr = mid * mid;

            if (midSqr == a) {
                return mid;     // the exact sqrt value
            } else if (midSqr < a) {
                start = mid;    // shift focus to bigger half
            } else {
                end = mid;      // shift focus to smaller half
            }
        }

        // if not find the exact sqrt value, return the approximated value with the defined precision
        return (start + end) / 2;
    }
}

Console Output

 

Count Occurance of a given number in sorted array

solution 1: linear search in O(N)

solution 2: binary search(preferred)

    need to add boundary checking for each sub-array

package org.fool.java.test;

public class NumberOccurrenceTest {
    public static void main(String[] args) {
        int[] nums = {1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 4, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 8, 9};
        System.out.println("Occurrence of num 2 is " + getOccurrence(2, nums, 0, nums.length - 1));
        System.out.println("Occurrence of num 6 is " + getOccurrence(6, nums, 0, nums.length - 1));
        System.out.println("Occurrence of num 7 is " + getOccurrence(7, nums, 0, nums.length - 1));
        System.out.println("Occurrence of num 17 is " + getOccurrence(17, nums, 0, nums.length - 1));
    }

    private static int getOccurrence(int k, int[] numbers, int startIndex, int endIndex) {
        if (endIndex < startIndex) {
            return 0;
        }

        if (numbers[startIndex] > k) {   // smallest element is larger than k
            return 0;
        }

        if (numbers[endIndex] < k) {     // largest element is smaller than k
            return 0;
        }

        if (numbers[startIndex] == k && numbers[endIndex] == k) {    // if all elements are k
            return endIndex - startIndex + 1;
        }

        int midIndex = (startIndex + endIndex) / 2;

        // a sub-array may possibly contain some other k
        if (numbers[midIndex] == k) {
            return 1 + getOccurrence(k, numbers, startIndex, midIndex - 1) + getOccurrence(k, numbers, midIndex + 1, endIndex);
        } else if (numbers[midIndex] > k) {  // only smaller half may contain k
            return getOccurrence(k, numbers, startIndex, midIndex - 1);
        } else {    // only bigger half may contain k
            return getOccurrence(k, numbers, midIndex + 1, endIndex);
        }
    }
}

Console Output

 

Find the 1st index of given number in a sorted array allowing duplicates

核心思想：二分查找

package org.fool.java.test;

public class FindFirstIndexTest {
    public static void main(String[] args) {
        int[] nums = {1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8, 8, 9};
        System.out.println("First index of number 1 is " + getFirstIndex(nums, 1, 0, nums.length - 1));
        System.out.println("First index of number 2 is " + getFirstIndex(nums, 2, 0, nums.length - 1));
        System.out.println("First index of number 3 is " + getFirstIndex(nums, 3, 0, nums.length - 1));
        System.out.println("First index of number 4 is " + getFirstIndex(nums, 4, 0, nums.length - 1));
        System.out.println("First index of number 5 is " + getFirstIndex(nums, 5, 0, nums.length - 1));
        System.out.println("First index of number 8 is " + getFirstIndex(nums, 8, 0, nums.length - 1));
    }

    // define a same header as a normal binary search
    // nums are the sorted array
    // a is the number we are looking for
    // start and end are the two index values to keep track of the current focus sub-array
    private static int getFirstIndex(int[] nums, int a, int start, int end) {
        if (end < start) {
            return -1;
        }

        if (nums[start] > a) {
            return -1;
        }

        if (nums[end] < a) {
            return -1;
        }

        // need to check if first element is a
        if (nums[start] == a) {
            return start;
        }

        int mid = (start + end) / 2;
        if (nums[mid] == a) {
            // we have 2 choices, either mid position is candidate, or the index we find in the left half
            int leftIndex = getFirstIndex(nums, a, start, mid - 1); // recursive call

            return leftIndex == -1 ? mid : leftIndex;
        } else if (nums[mid] > a) {  // which means a can only appear in the left half
            return getFirstIndex(nums, a, start, mid - 1);
        } else {    // which means only possible in the right half
            return getFirstIndex(nums, a, mid + 1, end);
        }
    }
}

Console Output

 

Reference

https://www.youtube.com/watch?v=XvsQ68jUc2U&index=50&list=PLlhDxqlV_-vkak9feCSrnjlrnzzzcopSG 

https://www.youtube.com/watch?v=W8923B642PQ&list=PLlhDxqlV_-vkak9feCSrnjlrnzzzcopSG&index=49 

https://www.youtube.com/watch?v=bvzJw9CVmkI&index=45&list=PLlhDxqlV_-vkak9feCSrnjlrnzzzcopSG