问题

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].

思路

简单的二分查找，由于有相同的元素，需要后面加了几个把附近的找出来。
ps：之前在公司项目也这样找过一次，不过要先排序再查找。

代码（越写越觉得不对劲，写的太冗余了，勉强ac）

class Solution {
    public int[] searchRange(int[] nums, int target) {
        //升序 找到起始的位置；原始的二分查找，再找旁边的几个。
        int[] find = new int[2];
        find[0]=-1;find[1]=-1;
        if(nums.length==0)
            return find;
        int result =binarySearch(0,nums.length-1,nums,target);
        if(result == -1)
            return find;
        else
           return findNearBy(result,target,nums);
            
            
    }
    public int binarySearch(int start,int end,int[] nums,int target){
        if(start>end){
            return -1;
        }
        int mid = (start+end)/2;
        if(nums[mid] == target){
            return mid;
        }
        if(target>nums[mid]){ //右边
            return  binarySearch(mid+1,end,nums,target);
        }else{
           return binarySearch(start,mid-1,nums,target);
        }
    }
    public int[] findNearBy(int index,int target,int[]nums){
        
        int start = index; int end =index;
        while(start>0){  //注意溢出  这里index=0没有算到，要后面判断一次
            if(nums[start]==target){
                start--;
            }else{
                break;
            }
        }
        // if(start ==0){   //0的情况
            if(nums[start]!=target){
                start = start+1;
            }
        // }
        while(end<nums.length-1){
            if(nums[end]==target){
                end ++;
            }else break;
        }
        // if(end == nums.length-1){  //length-1
            if(nums[end] != target){
                end = end-1;
            }
        // }
        int[] temp = new int[2];
        temp[0] = start;
        temp[1] = end;
        return temp;
        
    }
}