2.2 KiB
Leetcode Search-Insert-Position
2022-07-09 10:25
Algorithms:
#algorithm #binary_search
Data structures:
#DS #array
Difficulty:
#coding_problem #difficulty-easy
Additional tags:
#leetcode
Revisions:
N/A
Related topics:
tag:#binary_search
- Binary Search Algorithm
- Leetcode Binary-Search
- Leetcode First-Bad-Version
- Leetcode Lowest-Common-Ancestor-Of-a-Binary-Search-Tree
- Leetcode Search-a-2D-Matrix
- Leetcode Two-Sum-IV-Input-Is-a-BST
Links:
Problem
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.
Examples
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
numsare unique. numsis sorted in ascending order.
Thoughts
[!summary] This is a #binary_search problem.
similar to Leetcode First-Bad-Version, this relies on the converging nature of Bi-search
if the value is found, simply return it
if not found, at the end, position l - 1 should be smaller than val.
and position l should be bigger, which is the position for the answer:
If not, return the index where it would be if it were inserted in order.
Solution
Iteration
class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
// variant of bi-search
int r = nums.size() - 1;
int l = 0;
int mid, val;
do {
mid = l + (r - l) / 2;
val = nums[mid];
if (val == target) {
return mid;
} else if (val < target) {
l = mid + 1;
} else {
r = mid - 1;
}
} while (l <= r);
return l;
}
};