Search for a Range

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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]

public class Solution {    public int[] searchRange(int[] A, int target) {        int []res={-1,-1};        if(A.length<=0) return res;        res[0]=findLeft(A,target);        res[1]=findRight(A,target);        return res;    }    public int findLeft(int []A,int target){        int l=0;        int r=A.length-1;        while(l<=r){            int mid=(l+r)/2;            if(A[mid]>target){                r=mid-1;            }            else if(A[mid]<target){                l=mid+1;            }            else{                if(mid>0 && A[mid]==A[mid-1]){                    r=mid-1;                }                else return mid;            }        }        return -1;    }    public int findRight(int []A,int target){        int l=0;        int r=A.length-1;        while(l<=r){            int mid=(l+r)/2;            if(A[mid]>target){                r=mid-1;            }            else if(A[mid]<target){                l=mid+1;            }            else{                if(mid<A.length-1  && A[mid]==A[mid+1]){                    l=mid+1;                }                else return mid;            }        }        return -1;    }}

思路：二分查找

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