二分查找两种实现，附详细注释

输入: nums = [-1,0,3,5,9,12], target = 9
输出: 4
解释: 9 出现在 nums 中并且下标为 4

输入: nums = [-1,0,3,5,9,12], target = 2
输出: -1
解释: 2 不存在 nums 中因此返回 -1

class Solution(object):
    def search(self, nums, target):
        # 刚开始的取值区间：[0,len(nums) )
        left,right=0,len(nums)
        while right - left > 1:
            mid = (left+right) //2
            # nums[0]<=...<nums[mid] <= target 
            if nums[mid] <= target:
                # 遍历后区间改变为：[mid,right)
                left = mid
            else:
                # nums[right-1]...>=nums[mid+1]>=nums[mid] > target
                # 区间调整为：[left,right)
                right = mid
        # 迭代结束后
        # right-left=1，区间为[left, right=left+1)        
        # 即nums[left]就是二分后逼近的点
        # 判断nums[left]是否等于target即可
        return left if nums[left] == target else -1

class Solution(object):
    def search(self, nums, target):
        # 左右都是闭合区间的写法
        # 刚开始的取值区间：[0,len(nums)-1]
        left,right=0,len(nums)-1
        while left <= right:
            mid = (left+right) //2
            if nums[mid] == target:
                return mid
            # nums[0]<=...<nums[mid] < target 
            elif nums[mid] < target:
                # 遍历后区间改变为：[mid+1,right]
                left = mid + 1 
            else:
                # nums[right-1]...>=nums[mid+1]>=nums[mid] > target
                # 区间调整为：[left,mid-1]
                right = mid-1
        # 迭代结束后
        # left - right = 1，此时区间[left,right] = [left,left-1]变为空！
        # 所以返回 -1
        return -1