Leetcode-Search Insert Position

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.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

Have you met this question in a real interview?

 

Analysis:

It is a binary search problem. Just need to consider the corner cases carefully!

 

Newest solution:

We keep start and end as unchecked points, because we check mid, when update, we excluded mid. In this way, we will always converge to one point, i.e., start==end. In the next step, start will always point to the insert point.

With this solution, there is not corner case!

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

 

 

Solution:

public class Solution {
    public int searchInsert(int[] A, int target) {
        int len = A.length;
        if (len==0) return -1;
     
        int head = 0, end = A.length-1;
        
        //NOTE:Consider these coner cases  carefully!
        if (target<=A[head]) return head;
        if (target>A[end]) return end+1;
        else if (target==A[end]) return end;
        
        while (end-head>1){
            int mid = (head+end)/2;
            if (A[mid]==target) return mid;
            else if (A[mid]>target)
                end = mid;
            else
                head = mid;
        }
        return end;
        
    }
}