*binary Search

For a given sorted array (ascending order) and a number target , find the first index of this number in time O(log n) complexit Y.

If the target number does not exist in the array, return -1 .

Has you met this question in a real interview?

YesWhich company asked do you question?AirbnbAlibabaAmazonAppleBaiduBloombergCiscoDropboxEbayFacebook Google Hulu Intel Linkedin Microsoft NetEase Nvidia Oracle Pinterest Snapchat Tencent Twitter Uber Xiaomi Yahoo Yelp Zenefits Thanks for your feedback. Example

If the array is [1, 2, 3, 3, 4, 5, 10] , for given target 3 , return 2 .

Challenge

If The count of numbers is bigger than 2^32, can your code work properly?

1      Public intBinarySearch (int[] Nums,inttarget)2     {3         //Write your code here4         if(nums.length==0)return-1;5         intleft = 0;6         intright = Nums.length-1;7        While  (left+1 < right) 8         {9             intMid = (left + right)/2; Tenif(target==nums[mid])  right = mid; -  }  -              the             if(target<Nums[mid]) -             { -right = Mid-1; -             } +             if(target>Nums[mid]) -             { +left = mid+1; A             } at              -         } -          -         if(Nums[left]==target)returnLeft ; -         if(Nums[right]==target)returnRight ; -         return-1; in}

*binary Search