[leetcode]34. Search for a Range

题目：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].

分析：本身看起来很简单的题目，我到底是怎么写的这么复杂又冗余的。。。

代码：

class Solution {
public:
    int searchFirst(vector<int>& nums,int target,int low,int high){
        if(nums[high]<target) return -1;
        while(low<=high){
            if(nums[low]==target) return low;
            int mid=low+(high-low)/2;
            if(mid==low) {
                if(nums[low]==target){
                    return low;
                }
                else{
                    if(nums[high]==target){
                        return high;
                    }
                    else{
                        return -1;
                    }
                }
            }
            if(nums[mid]>=target){
                high=mid;
            }
            else{
                low=mid;
            }
        }
        return -1;
    }
    int searchLast(vector<int>& nums,int target,int low,int high){
        if(nums[low]>target) return -1;
        while(low<=high){
            if(nums[high]==target) return high;
            int mid=low+(high-low)/2;
            if(mid==low){
                if(nums[high]==target){
                    return high;
                }
                else{
                    if(nums[low]==target){
                        return low;
                    }
                    else{
                        return -1;
                    }
                }
            }
            if(nums[mid]<=target){
                low=mid;
            }
            else{
                high=mid;
            }
        }
        return -1;
    }
    vector<int> searchRange(vector<int>& nums, int target) {
        int n=nums.size();
        vector<int> ret(2,-1);
        if(n==0) return ret;
        ret[0]=(searchFirst(nums,target,0,n-1));
        ret[1]=(searchLast(nums,target,0,n-1));
        return ret;
    }
};

附上第一次刷题时的代码，然而这个运行时间好像更慢一些：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int m=nums.size();
        vector<int> result(2,-1);
        int left=0;
        int right=m-1;
        while(left<right-1){
            int mid=left+(right-left)/2;
            if(nums[mid]==target){
                right=mid;
            }
            else if(nums[mid]>target){
                right=mid;
            }
            else{
                left=mid;
            }
            
        }
        if(nums[left]==target){
            result[0]=left;
        }
        else if(nums[right]==target){
            result[0]=right;
        }
        else{
            return result;
        }
        left=0;
        right=m-1;
        while(left<right-1){
            int mid=left+(right-left)/2;
            if(nums[mid]==target){
                left=mid;
            }
            else if(nums[mid]>target){
                right=mid;
            }
            else{
                left=mid;
            }
        }
        if(nums[right]==target){
            result[1]=right;
        }
        else if(nums[left]==target){
            result[1]=left;
        }
        else{
            return result;
        }
        return result;
    }
};