Problem
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 (Java)
class Solution {
public int searchInsert(int[] nums, int target) {
int low = 0, high = nums.length;
while(low < high) {
// Calculate middle index
int mid = low + (high - low)/2;
if(target > nums[mid])
low = mid + 1;
else
high = mid;
}
return low;
}
}
Solution (Javascript)
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var searchInsert = function(nums, target) {
var len = nums.length;
var left = 0;
var right = len - 1;
var mid = 0;
if (!len) return 0;
while (left <= right) {
mid = Math.floor((left + right) / 2);
if (nums[mid] > target) {
right = mid - 1;
} else if (nums[mid] < target) {
left = mid + 1;
} else {
return mid;
}
}
return (nums[mid] > target) ? mid : (mid + 1);
};
Explain:
nope.
Complexity:
- Time complexity : O(log(n)).
- Space complexity : O(1).