[LeetCode] #34 Search for a Range

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].

本题利用二分算法的思想，寻找target的开始和结束位置。时间：17ms。代码如下：

class Solution {
public:
    int searchfRange(vector<int>& nums, int f, int l, int target){
        if (f > l || nums[l] != target)
            return l + 1;
        if (nums[f] == target)
            return f;
        int mid = f + (l - f) / 2;
        if (nums[mid] == target)
            return searchfRange(nums, f, mid - 1, target);
        else
            return searchfRange(nums, mid + 1, l, target);
    }
    int searchlRange(vector<int>& nums, int f, int l, int target){
        if (f > l || nums[f] != target)
            return f - 1;
        if (nums[l] == target)
            return l;
        int mid = f + (l - f) / 2;
        if (nums[mid] == target)
            return searchlRange(nums, mid + 1, l, target);
        else
            return searchlRange(nums, f, mid - 1, target);
    }
    vector<int> searchRange(vector<int>& nums, int f, int l, int target){
        if (f > l || target < nums[f] || target > nums[l])
            return vector<int>(2, -1);
        vector<int> range(2,-1);
        int mid = f + (l - f) / 2;
        if (nums[f] == target){
            range[0] = f;
            range[1] = searchlRange(nums, f+1, l, target);
            return range;
        }
        if (nums[l] == target){
            range[0] = searchfRange(nums, f, l-1, target);
            range[1] = l;
            return range;
        }
        if (nums[mid] == target){
            range[0] = searchfRange(nums, f, mid - 1, target);
            range[1] = searchlRange(nums, mid + 1, l, target);
            return range;
        }
        else if (nums[mid] < target)
            return searchRange(nums, mid + 1, l, target);
        else
            return searchRange(nums, f, mid - 1, target);
    }
    vector<int> searchRange(vector<int>& nums, int target) {
        if (target<nums.front() || target>nums.back())
            return vector<int>(2, -1);
        return searchRange(nums, 0, nums.size() - 1, target);
    }
};