34. Search for a Range

原题

Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,

Given [5, 7, 7, 8, 8, 10] and target value 8,return [3, 4].

自己做题总结出的几个例子

        [1,6] target=3 return [-1,-1]        [1,6] target=6, returns [1,1]        [1,6,6,6] target=6 returns[1,3]

代码实现

这是一个O(Logn)时间复杂度的算法

        public int[] SearchRange(int[] nums, int target)        {            int[] rtn = new int[2];            int lo = 0, hi = nums.Length;            //after finishing this loop, we can get the right boundary of range for target             while (lo < hi){                int mid = lo + (hi - lo) / 2;                if (nums[mid] <= target) lo = mid + 1;                else hi = mid;            }            rtn[1] = lo-1;            //no existence: three conditions            if (rtn[1] == -1 || rtn[1] == nums.Length ||                 (rtn[1] >= 0 && nums[rtn[1]] != target)) {                rtn[0] = -1;                rtn[1] = -1;                return rtn;            }            //following to solve the series of values with equal to target,            //so get the left range for target            lo = 0;            while (lo < hi){                int mid = lo + (hi - lo) / 2;                if (nums[mid] >= target) hi = mid;                else {                    lo = mid + 1;                    if (nums[lo] == target){                        rtn[0] = lo;break;                    }                }            }            return rtn;        }