Search for a Range

描述

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

已经排好了序,用二分查找。

重新实现 lower_bound 和 upper_bound

// Search for a Range
// 重新实现 lower_bound 和 upper_bound
// 时间复杂度O(logn),空间复杂度O(1)
class Solution {
public:
    vector<int> searchRange (vector<int>& nums, int target) {
        auto lower = lower_bound(nums.begin(), nums.end(), target);
        auto uppper = upper_bound(lower, nums.end(), target);

        if (lower == nums.end() || *lower != target)
            return vector<int> { -1, -1 };
        else
            return vector<int> {distance(nums.begin(), lower), distance(nums.begin(), prev(uppper))};
    }

    template<typename ForwardIterator, typename T>
    ForwardIterator lower_bound (ForwardIterator first,
            ForwardIterator last, T value) {
        while (first != last) {
            auto mid = next(first, distance(first, last) / 2);

            if (value > *mid) first = ++mid;
            else              last = mid;
        }

        return first;
    }

    template<typename ForwardIterator, typename T>
    ForwardIterator upper_bound (ForwardIterator first,
            ForwardIterator last, T value) {
        while (first != last) {
            auto mid = next(first, distance (first, last) / 2);

            if (value >= *mid) first = ++mid;  // 与 lower_bound 仅此不同
            else               last = mid;
        }

        return first;
    }
};

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