Description
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] <= 104nums contains distinct values sorted in ascending order.-104 <= target <= 104
Solutions
Solution 1: Binary Search
Since the array nums is already sorted, we can use the binary search method to find the insertion position of the target value target.
The time complexity is O(log n), and the space complexity is O(1). Here, n is the length of the array nums.
PythonJavaC++GoTypeScriptRustJavaScriptPHP
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
l, r = 0, len(nums)
while l < r:
mid = (l + r) >> 1
if nums[mid] >= target:
r = mid
else:
l = mid + 1
return l(code-box)
class Solution {
public int searchInsert(int[] nums, int target) {
int l = 0, r = nums.length;
while (l < r) {
int mid = (l + r) >>> 1;
if (nums[mid] >= target) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}(code-box)
class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int l = 0, r = nums.size();
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] >= target) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
};(code-box)
func searchInsert(nums []int, target int) int {
l, r := 0, len(nums)
for l < r {
mid := (l + r) >> 1
if nums[mid] >= target {
r = mid
} else {
l = mid + 1
}
}
return l
}(code-box)
function searchInsert(nums: number[], target: number): number {
let [l, r] = [0, nums.length];
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] >= target) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}(code-box)
impl Solution {
pub fn search_insert(nums: Vec<i32>, target: i32) -> i32 {
let mut l: usize = 0;
let mut r: usize = nums.len();
while l < r {
let mid = (l + r) >> 1;
if nums[mid] >= target {
r = mid;
} else {
l = mid + 1;
}
}
l as i32
}
}(code-box)
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var searchInsert = function (nums, target) {
let [l, r] = [0, nums.length];
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] >= target) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};(code-box)
class Solution {
/**
* @param Integer[] $nums
* @param Integer $target
* @return Integer
*/
function searchInsert($nums, $target) {
$l = 0;
$r = count($nums);
while ($l < $r) {
$mid = $l + $r >> 1;
if ($nums[$mid] >= $target) {
$r = $mid;
} else {
$l = $mid + 1;
}
}
return $l;
}
}(code-box)
Solution 2: Binary Search (Built-in Function)
We can also directly use the built-in function for binary search.
The time complexity is O(log n), where n is the length of the array nums. The space complexity is O(1).
PythonJavaC++Go
class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
return bisect_left(nums, target)(code-box)
class Solution {
public int searchInsert(int[] nums, int target) {
int i = Arrays.binarySearch(nums, target);
return i < 0 ? -i - 1 : i;
}
}(code-box)
class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
return lower_bound(nums.begin(), nums.end(), target) - nums.begin();
}
};(code-box)
func searchInsert(nums []int, target int) int {
return sort.SearchInts(nums, target)
}(code-box)
