二分查找——边界问题（leetcode 34，Offer 53

"""
寻找决策边界分为寻找左侧边界、右侧边界
寻找左侧边界，压缩右边界（当mid=target缩小right）
寻找右侧边界，压缩左边界（当mid=target缩小left）
"""

#寻找左侧边界
def query_left_target_boundary_v1(list_num, target):
    left = 0
    right = len(list_num) - 1
    while left <= right:
        mid = left + int((right-left)/2)
        if list_num[mid] > target:
            right = mid - 1
        elif list_num[mid] < target:
            left = mid + 1
        elif list_num[mid] == target:
            right = mid - 1 #缩小right,逼近左侧边界
    if left >= len(list_num) or list_num[left] != target: #针对特殊的左侧边界，进行特殊的判断
        return -1
    return left

def query_left_target_boundary_v2(list_num, target):
    left = 0
    right = len(list_num)
    while left < right:
        mid = left + int((right-left)/2)
        if list_num[mid] > target:
            right = mid
        elif list_num[mid] < target:
            left = mid + 1
        elif list_num[mid] == target:
            right = mid
    if left >= len(list_num) or list_num[left] != target:
        return -1
    return left

#寻找右侧边界
def query_right_target_boundary_v1(list_num, target):
    left = 0
    right = len(list_num) - 1
    while left <= right:
        mid = left + int((right-left)/2)
        if list_num[mid] > target:
            right = mid - 1
        elif list_num[mid] < target:
            left = mid + 1
        elif list_num[mid] == target:
            left = mid + 1 #缩小left,逼近右侧侧边界
    if right < 0 or list_num[right] != target: #针对特殊的右侧边界，进行特殊的判断
        return -1
    return right

def query_left_target_boundary_v2(list_num, target):
    left = 0
    right = len(list_num)
    while left < right:
        mid = left + int((right-left)/2)
        if list_num[mid] > target:
            right = mid
        elif list_num[mid] < target:
            left = mid + 1
        elif list_num[mid] == target:
            left = mid + 1
    if right <= 0 or list_num[right - 1] != target:
        return -1
    return right - 1

"""
剑指 Offer 53 - I. 在排序数组中查找数字 I
统计一个数字在排序数组中出现的次数。

示例 1:
输入: nums = [5,7,7,8,8,10], target = 8
输出: 2

示例 2:
输入: nums = [5,7,7,8,8,10], target = 6
输出: 0

"""
给定一个按照升序排列的整数数组nums, 和一个目标值target。找出给定目标值在数组的开始位置和结束位置(leetcode 34)
"""

"""
解题思路：这道题和find_target_number类似，也是考虑求出目标值的左右边界
"""
def find_target_first_end(nums, target):
    #找出左侧边界的索引
    left_index = find_left_index(nums, target)
    if left_index == -1:
        return [-1, -1]
    right_index = find_right_index(nums, target)
    return [left_index, right_index]

def find_left_index(nums, target):
    left = 0
    right = len(nums)
    while left < right:
        mid = left + int((right - left) / 2)
        if nums[mid] >= target:
            right = mid
        else:
            left = mid + 1
    if left >= len(nums) or nums[left] != target:
        return -1
    return left

def find_right_index(nums, target):
    left = 0
    right = len(nums)
    while left < right:
        mid = left + int((right - left) / 2)
        if nums[mid] <= target:
            left = mid + 1
        else:
            right = mid
    if right <= 0 or nums[right - 1] != target:
        return -1
    return right - 1

if __name__ == "__main__":
    nums = [5,7,7,8,8,10]
    target = 10
    res = find_target_first_end(nums, target)
    print(res)

解题思路：
按照二分查找的方式，找到目标值的左右边界的索引 然后用右边界-左边界+1
"""
class Solution(object):
    def search(self, nums, target):
        left = 0
        right = len(nums)
        target_left = -1
        target_right = -1
        #找出目标值的左侧边界
        while left < right:
            mid = left + int((right - left) / 2)
            if nums[mid] > target:
                right = mid
            elif nums[mid] < target:
                left = mid + 1
            elif nums[mid] == target:
                right = mid

        if left >= len(nums) or nums[left] != target:
            target_left = -1
        else:
            target_left = left