算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## 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

{% if book.java %}
{% codesnippet "./code/search-for-a-range."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// 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 searchRange (vector& 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 { -1, -1 };
else
return vector {distance(nums.begin(), lower), distance(nums.begin(), prev(uppper))};
}

template
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
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;
}
};
```
{% endif %}


### 相关题目

* [Search Insert Position](search-insert-position.md)