Search for a Range

Posted 2020-06-13

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].

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res;
        res.push_back(bsarch_left(nums, target));
        res.push_back(bsarch_right(nums, target));
        return res;
    }

    int bsarch_left(vector<int>& nums, int target) {
        int i = 0;
        int j = nums.size() - 1;
        while (i < j) {
            int mid = i + (j-i) / 2;
            if (nums[mid] < target) {
                i = mid + 1;
            } else {
                j = mid;
            }
        }

        if (target == nums[i]) {
            return i;
        }
        else {
            return -1;
        }
    }

    int bsarch_right(vector<int>& nums, int target) {
        int i = 0;
        int j = nums.size() - 1;
        while (i < j) {
            int mid = i + (j-i + 1) / 2;
            if (nums[mid] <= target) {
                i = mid;
            } else {
                j = mid - 1;
            }
        }

        if (target == nums[i]) {
            return i;
        }
        else {
            return -1;
        }
    }
};