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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.

Example 1:

Input: nums = [1,3,5,6], target = 5 Output: 2 Example 2:

Input: nums = [1,3,5,6], target = 2 Output: 1 Example 3:

Input: nums = [1,3,5,6], target = 7 Output: 4

Constraints:

1 <= nums.length <= 104 -104 <= nums[i] <= 104 nums contains distinct values sorted in ascending order. -104 <= target <= 104

Solution:

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.

 

Example 1:

Input: nums = [1,3,5,6], target = 5
Output: 2
Example 2:

Input: nums = [1,3,5,6], target = 2
Output: 1
Example 3:

Input: nums = [1,3,5,6], target = 7
Output: 4
 

Constraints:

1 <= nums.length <= 104
-104 <= nums[i] <= 104
nums contains distinct values sorted in ascending order.
-104 <= target <= 104

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

}

};

优化

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

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