LeetCode34：Find First and Last Position of Element in Sorted Array(二分法)

Given an array of integers nums 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 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

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LeetCode：链接

首先要弄懂自己写的二分法：链接

方法一：使用二分法找到目标值，然后以此目标值为中心，向左右两边扩展，但要考虑边界的情况。或是找到nums[mid]==target，然后从区间[low、high]的边缘向内缩小，来确定范围。

方法二(最优解)：同样使用二分法，先找到左端点，然后继续使用二分法去探查右端点。

一定要注意，查找左端点返回的是左指针，查找右端点返回的是右指针！

如果找不到，比如12没有，最终end是9， start是10；比如0没有，最终start是0， end是-1！所有最终二分法的start和end都回归到这种结果！！！！

class Solution:
    def searchRange(self, nums, target):
        def binarySearchLeft(nums, target):
            low, high = 0, len(nums) - 1
            while low <= high:
                mid = (low + high) // 2
                if target > nums[mid]: 
                    low = mid + 1
                else: 
                    high = mid - 1
            # 左边界返回的是low
            return low

        def binarySearchRight(nums, target):
            low, high = 0, len(nums) - 1
            while low <= high:
                mid = (low + high) // 2
                if target >= nums[mid]: 
                    low = mid + 1
                else: 
                    high = mid - 1
            # 找右边界返回的是high
            return high
            
        left, right = binarySearchLeft(nums, target), binarySearchRight(nums, target)
        return (left, right) if left <= right else [-1, -1]