Search for a Range

LeetCode #34

Description:

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].

Example:

Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Idea:

重新实现lowerBound() and upperBound().

此处upperBound返回的值需要回走一步，因为upperBound返回的并不是要找的值。

Code:

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        if(nums.empty()) return vector<int> { -1, -1 };

        int left = findLowerBound(nums, target);
        int right = findUpperBound(nums, target) - 1;

        if(left == nums.size() || nums[left] != target)
            return vector<int>{-1, -1};
        else
            return vector<int>{left, right};
    }

    // 实现 lowerBound() 
    int findLowerBound(vector<int>& nums, int target){
        int mid, start, end;
        start=0;
        end=nums.size();

        while(start != end){
            mid = start + (end - start)/2;
            if(nums[mid] < target) start = mid + 1;
            else end=mid;
        }
        return start;
    }

    // 实现 upperBound()
    int findUpperBound(vector<int>& nums, int target){
        int mid, start, end;
        start=0;
        end=nums.size();

        while(start != end){
            mid = start + (end - start)/2;
            if(nums[mid] <= target) start = mid + 1;
            else end=mid;
        }
        return start;
    }
};