算法分析與設計課程（5）：【leetcode】Search for a Range

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].

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

算法分析：

1. 二分法

2. 二分法的框架, 需要考慮的位置有 3 個, 在代碼中我標了出來, 分別爲 q1, q2, q3

3. q1 是取 <= 還是取 <. 我的經驗是, 若是題目要求找到 target, 那麼就用 <=, 否則用 <. 我記得在二分搜索題時, 都是用 < 的

4. q2 比較容易, 考慮當 low == high 時, 我們希望遊標往哪裏走

5. q3, 返回 low/high. q3 的選取與 q2 有關. 還是需要考慮當 low == high 時, 遊標會往哪走

代碼：

class Solution:
    def searchRange(self, A, target):
        if len(A) == 0:
            return [-1, -1]

        start, end = 0, len(A) - 1
        while start + 1 < end:
            mid = (start + end) / 2
            if A[mid] < target:
                start = mid
            else:
                end = mid

        if A[start] == target:
            leftBound = start
        elif A[end] == target:
            leftBound = end
        else:
            return [-1, -1]

        start, end = leftBound, len(A) - 1
        while start + 1 < end:
            mid = (start + end) / 2
            if A[mid] <= target:
                start = mid
            else:
                end = mid
        if A[end] == target:
            rightBound = end
        else:
            rightBound = start
        return [leftBound, rightBound]