Python binary search detailed explanation-target value and left and right boundary unified template

Algorithm cheat sheet from labuladong 

Given an ascending sequence, find 3 cases: 
1. Find a value 
2. Find the left boundary 
3. Find the unified routine of the right boundary , use the "search interval" with both ends closed , just pay attention to nums[mid] == The code at the target condition and finally determine whether the left and right indexes are out of bounds

class Solution:
    def binary_search(self, nums, target):
        """
        找不到 target 返回 -1
        :param nums:
        :param target:
        :return:
        """
        if not nums:
            return -1

        left = 0
        right = len(nums) - 1

        # 搜索区间 左右闭
        while left <= right:
            mid = left + (right - left) // 2  # 地板除，向下取整
            if nums[mid] > target:
                # mid 大 => 向左侧区间查找
                right = mid - 1
            elif nums[mid] < target:
                # mid 小 => 向右侧区间查找
                left = mid + 1
            elif nums[mid] == target:
                # 找到 target 直接返回
                return mid
        # 搜索区间为空，没有target
        return -1

    def search_left_bound(self, nums, target):
        if not nums:
            return -1
        left = 0
        right = len(nums) - 1

        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] > target:
                # mid大 => 向左侧区间查找
                right = mid - 1
            elif nums[mid] < target:
                # mid小 => 向右侧区间查找
                left = mid + 1
            elif nums[mid] == target:
                # 此时说明，左侧边界只能小于等于mid => 左侧区间
                right = mid - 1

        # 最后，判断left是否越界，如果 target 比所有数都大，l 会达到nums的size
        if left >= len(nums) or nums[left] != target:
            return -1
        return left

    def search_right_bound(self, nums, target):
        if not nums:
            return -1
        left = 0
        right = len(nums) - 1

        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] > target:
                right = mid - 1
            elif nums[mid] < target:
                left = mid + 1
            elif nums[mid] == target:
                # 此时说明，右侧边界只能大于等于mid => 右侧区间
                left = mid + 1

        # 检查 right 是否越界
        if right < 0 or nums[right] != target:
            return -1
        return right