Classic Binary Search

Description

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.

Code

Time Complexity: $O(\log n)$, Space Complexity: $O(1)$

基本概念同 Binary Search 101。

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while(left < right) {
            int mid = left + (right - left) / 2;
            if(nums[mid] >= target) right = mid;
            else left = mid + 1;
        }
 
        return nums[left] == target ? left : -1;
    }
};

Source

• Binary Search - LeetCode