LeetCode-in-All
35. Search Insert Position
Easy
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
Example 4:
Input: nums = [1,3,5,6], target = 0
Output: 0
Example 5:
Input: nums = [1], target = 0
Output: 0
Constraints:
1 <= nums.length <= 104-104 <= nums[i] <= 104numscontains distinct values sorted in ascending order.-104 <= target <= 104
Solution
#include <stdio.h>
int searchInsert(int* nums, int numsSize, int target) {
int lo = 0;
int hi = numsSize - 1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (target == nums[mid]) {
return mid;
} else if (target < nums[mid]) {
hi = mid - 1;
} else { // target > nums[mid]
lo = mid + 1;
}
}
return lo; // return the insertion point
}