Clean short Java code solution log(N) binary search


  • 0
    I

    Simple Java binary search implementation in log(n):

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

Log in to reply
 

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.