Search for a Range ——LeetCode

Given a sorted array of integers, 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].

 

题目大意：给定一个排序好的数组，找到指定数的起始范围，如果不存在返回[-1,-1]；

解题思路：要求O(lgN)时间复杂度，二分查找。

    public int[] searchRange(int[] nums, int target) {

        int[] res = new int[2];

        if(nums==null||nums.length==0){

            return res;

        }

        res[0]=getLow(nums,target,0,nums.length-1);

        res[1]=getHigh(nums,target,0,nums.length-1);

        return res;

    }

    int getLow(int[] nums,int target,int low,int high){

        int mid=(low+high)>>1;

        if(low>high){

            return -1;

        }

        if(low==high){

            return nums[low]==target?low:-1;

        }

        if(nums[mid]==target){

            return getLow(nums,target,low,mid);

        }

        if(nums[mid]<target){

            low=mid+1;

            return getLow(nums,target,low,high);

        }else{

            high=mid-1;

            return getLow(nums,target,low,high);

        }

    }

    int getHigh(int[] nums,int target,int low,int high){

         int mid=(low+high)>>1;

         if(low>high){

             return -1;

         }

        if(low==high){

            return nums[low]==target?low:-1;

        }

        if(nums[mid]==target){

            int tmp=getHigh(nums,target,mid+1,high);

            int max=Math.max(tmp,mid);

            return max;

        }

        if(nums[mid]<target){

            low=mid+1;

            return getHigh(nums,target,low,high);

        }else{

            high=mid-1;

            return getHigh(nums,target,low,high);

        }

    }