Test1

由bqjdzn7f创建，最终由bqjdzn7f更新于2024-05-20 02:09 被浏览 94 用户

#102
def int_most(arr):
    max_element = None
    max_count = 0
    current_element = None
    current_count = 0

    for element in arr:
        if element == current_element:
            current_count += 1
        else:
            current_element = element
            current_count = 1

        if current_count > max_count:
            max_count = current_count
            max_element = current_element

    return max_element

# 示例用法
sorted_array = [1, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5]
result = int_most(sorted_array)
print("出现次数最多的元素是:", result)
#103
def re_duplicates(nums):
    if not nums:
        return 0

    unique_count = 1  # 用于记录不重复元素的个数

    for i in range(1, len(nums)):
        if nums[i] != nums[i - 1]:
            nums[unique_count] = nums[i]
            unique_count += 1

    return unique_count

# 示例用法
sorted_array = [1, 1, 2, 2, 2, 3, 4, 4, 5]
new_length = re_duplicates(sorted_array)
print("去重后的数组：", sorted_array[:new_length])
#104
def find_two_sum_count(nums, target):
    count = 0  # 用于记录符合条件的情况数
    left, right = 0, len(nums) - 1

    while left < right:
        current_sum = nums[left] + nums[right]

        if current_sum == target:
            count += 1
            left += 1  # 移动左指针，寻找下一个可能的组合
            right -= 1  # 移动右指针，寻找下一个可能的组合
        elif current_sum < target:
            left += 1  # 和太小，增加左边的数
        else:
            right -= 1  # 和太大，减小右边的数

    return count

# 示例用法
sorted_array = [1, 2, 3, 4, 5, 6, 7, 8, 9]
target = 10
count = find_two_sum_count(sorted_array, target)
print("符合条件的情况数为:", count)

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