# Search Insert Position

Easy

Given a sorted array 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 may assume no duplicates in the array.

Example 1:

``````Input:
[1,3,5,6], 5

Output:
2
``````

Example 2:

``````Input:
[1,3,5,6], 2

Output:
1
``````

Example 3:

``````Input:
[1,3,5,6], 7

Output:
4
``````

Example 4:

``````Input:
[1,3,5,6], 0

Output:
0
``````

## Solution

Template #1

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

return left;
}
}
``````

Template #2

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

return left;
}
}
``````

Template #3

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

if (target > nums[right]) {
return right + 1;
}
if (target > nums[left]) {
return left + 1;
}
return left;
}
}
``````