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].

Solution:

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

C语言源代码（8ms）：

int* searchRange(int* nums, int numsSize, int target, int* returnSize) {
    int* res=(int*)malloc(sizeof(int)*2);
    int l=0,r=numsSize-1,mid;
    while(l<=r){
        mid=(l+r)>>1;
        if(nums[mid]==target)break;
        else if(nums[mid]>target)r=mid-1;
        else l=mid+1;
    }
    if(l<=r){
        l=mid-1;
        while(l>=0 && nums[l]==nums[mid])l--;
        r=mid+1;
        while(r<numsSize && nums[r]==nums[mid])r++;
        res[0]=l+1;res[1]=r-1;
        *returnSize=2;
        return res;
    }else{
        res[0]=-1;res[1]=-1;
        *returnSize=2;
        return res;
    }
}

C++源代码（14ms）：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res;
        int l=0,len=nums.size(),r=len-1,mid;
        while(l<=r){
            mid=(l+r)>>1;
            if(nums[mid]==target)break;
            else if(nums[mid]>target)r=mid-1;
            else l=mid+1;
        }
        if(l<=r){
            l=mid-1;
            while(l>=0 && nums[l]==nums[mid])l--;
            r=mid+1;
            while(r<len && nums[r]==nums[mid])r++;
            res.push_back(l+1);
            res.push_back(r-1);
        }else{
            res.push_back(-1);
            res.push_back(-1);
        }
        return res;
    }
};

Python源代码（58ms）：

class Solution:
    # @param {integer[]} nums
    # @param {integer} target
    # @return {integer[]}
    def searchRange(self, nums, target):
        length=len(nums);l=0;r=length-1;mid=0
        res=[-1,-1]
        while l<=r:
            mid=(l+r)>>1;
            if nums[mid]==target:break
            elif nums[mid]>target:r=mid-1
            else:l=mid+1
        if l<=r:
            l=mid-1
            while l>=0 and nums[l]==nums[mid]:l-=1
            r=mid+1
            while r<length and nums[r]==nums[mid]:r+=1
            res[0]=l+1;res[1]=r-1
        return res