Search Insert Position Solution

Question

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would is if it were inserted in order.

Assume no duplicates in the array.

Here is 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

Solution 1--Naive

Iterate the array and compare target with ith and (i+1) th element. Time complexity O (n).

1  Public classSolution {2      Public intSearchinsert (int[] Nums,inttarget) {3         if(nums==NULL)return0;4         if(Target <= nums[0])return0;5          for(inti=0; i<nums.length-1; i++){6             if(Target > Nums[i] && target <= nums[i+1]){7                 returnI+1;8             }9         }Ten         returnnums.length; One     } A}

Solution 2--Binary Search

If the target number doesn ' t exist in original array and then after iteration, it must is pointed by low pointer.

Time Complexity O (log (n))

1  Public classSolution {2      Public intSearchinsert (int[] Nums,inttarget) {3         if(Nums = =NULL|| Nums.length = = 0)4             return0;5         intStart = 0, end = nums.length-1, Mid = (End-start)/2 +start;6          while(Start <=end) {7Mid = (End-start)/2 +start;8             if(Nums[mid] = =target)9                 returnmid;Ten             Else if(Nums[mid] <target) OneStart = mid + 1; A             Else -End = Mid-1; -         } the         returnstart; -     } -}

Search Insert Position Solution