二分查找——leetcode 209 长度最小的连续子数组

"""
leetcode 209
求长度最小的连续子数组 target = 7, nums = [2,3,1,2,4,3] output=2

解题思路：
这个题可以采用滑动窗口的方法，他是窗口大小变化的一种窗口计算方法[在后续有单独的滑动窗口模块]，时间复杂度为O(n)；同时，这道题也可以采用二分查找方法
"""

#基于滑动窗口的方法
def search_min_length_array(nums, target):
    left = 0
    right = 0
    ans = float('inf')
    total = 0
    while right < len(nums):
        total += nums[right]
        right += 1
        while total >= target:
            if total == target:
                ans = min(ans, right - left)
            total -= nums[left]
            left += 1
    return 0 if ans == float('inf') else ans


#基于二分查找的方法
def search_min_length_array_II(nums, target):
    sums = [0]
    for num in nums:
        sums.append(sums[-1] + num)
    #找到大于目标值得决策边界
    right_index = search_right_index(sums, target)
    #找到相差一个目标值的索引
    ans = float('inf')
    for index in range(right_index, len(sums)):
        target2 = sums[index] - target
        left_index = search_left_index(sums, target2, index)
        if left_index == -1:
            continue
        else:
            ans = min(ans, index - left_index)
    return 0 if ans == float('inf') else ans


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


def search_left_index(sums, target, right_index):
    left = 0
    right = right_index
    while left < right:
        mid = left + int((right - left) / 2)
        if sums[mid] == target:
            return mid
        elif sums[mid] > target:
            right = mid
        elif sums[mid] < target:
            left = mid + 1

    return -1



if __name__ == "__main__":
    nums = [1,1,1,1,1,1,1,1]
    target = 11
    res = search_min_length_array_II(nums, target)
    print(res)