704

704. Binary Search (Easy)

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

 

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

 

Constraints:

• 1 <= nums.length <= 104
• -104 < nums[i], target < 104
• All the integers in nums are unique.
• nums is sorted in ascending order.

Companies:
Amazon, Apple, Adobe, Google, SAP, Bloomberg, Facebook, Microsoft, tcs

Related Topics:
Array, Binary Search

Similar Questions:

• Search in a Sorted Array of Unknown Size (Medium)

Solution 1. L <= R

// OJ: https://leetcode.com/problems/binary-search/
// Author: github.com/lzl124631x
// Time: O(logN)
// Space: O(1)
class Solution {
public:
    int search(vector<int>& A, int target) {
        int L = 0, R = A.size() - 1;
        while (L <= R) {
            int M = (L + R) / 2;
            if (A[M] == target) return M;
            if (A[M] < target) L = M + 1;
            else R = M - 1;
        }
        return -1;
    }
};

Solution 2. L < R

// OJ: https://leetcode.com/problems/binary-search/
// Author: github.com/lzl124631x
// Time: O(logN)
// Space: O(1)
class Solution {
public:
    int search(vector<int>& A, int target) {
        int L = 0, R = A.size() - 1;
        while (L < R) {
            int M = (L + R) / 2;
            if (A[M] < target) L = M + 1;
            else R = M;
        }
        return A[L] == target ? L : -1;
    }
};