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Easy Binary Search

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2020-10-08

Question:

Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1.

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

Note:

  1. You may assume that all elements in nums are unique.
  2. n will be in the range [1, 10000].
  3. The value of each element in nums will be in the range [-9999, 9999].

Solution:

This will be a template for binary search

class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int start = 0;
        int end = nums.length - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                return mid;        
            } else if (nums[mid] > target) {
                end = mid;
            } else {
                start = mid;
            }
        }
        
        if (nums[start] == target) {
            return start;
        }
        
        if (nums[end] == target) {
            return end;
        }
        
        return -1;

    }
}

Using start + 1 < end will avoid the infinite loop problem as following. If we are having [1,1], it will cause infinite loop.

while (start <= end) {
	    mid = (start + end) / 2;
			if (A[mid] <= target) {
			    start = mid; 
			} else {
			    end = mid - 1;
    }
}