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Binary Search - Example

Binary Search - Example


Leetcode 35 Search Insert Position

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

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

Constraints:

1 <= nums.length <= 10^4

-10^4 <= nums[i] <= 10^4

nums contains distinct values sorted in ascending order.

-10^4 <= target <= 10^4


Analysis:

The array is sorted, so we just need to run a routine binary search. Make a guest first, then based on the guessed result, we can adjust the search range.


See the code below:


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


If you know the lower_bound() function in C++, then the code could become quite concise:

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        return lower_bound(nums.begin(), nums.end(), target) - nums.begin();
    }
};



Upper Layer


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