UIC All QUIZ 类似题目-模拟考

def count_word_in_sentence(sentence, word):
    # Split the sentence into words
    words = sentence.split()

    # Initialize counter for the word
    count = 0

    # Check each word in the sentence
    for w in words:
        if w == word:
            count += 1

    return count

sentence = "Artificial intelligence is the future. Intelligence can change the world."
word = "intelligence"
print(count_word_in_sentence(sentence, word))  # Expected output: 1

def find_pattern(text, pattern):
    text = text.lower()
    pattern = pattern.lower()
    indices = []
    index = -1  # Start at -1 so the first run adds 0 to start the search
    while True:
        # Find the next occurrence of the pattern
        index = text.find(pattern, index + 1)
        # If no more occurrences, break the loop
        if index == -1:
            break
        # Otherwise, append the index to the list
        indices.append(index)
    return indices

text = "The quick brown fox jumps over the lazy dog. The dog is not that lazy after all."
pattern = "the"
print(find_pattern(text, pattern))

def find_pattern(text, pattern):
    text = text.lower()
    pattern = pattern.lower()
    indices = []
    index = -1  # Start at -1 so the first run adds 0 to start the search
    while True:
        # Find the next occurrence of the pattern
        index = text.find(pattern, index + 1)
        # If no more occurrences, break the loop
        if index == -1:
            break
        # Otherwise, append the index to the list
        indices.append(index)
    return indices

text = "The quick brown fox jumps over the lazy dog. The dog is not that lazy after all."
pattern = "the"
print(find_pattern(text, pattern))

def longest_substring_without_repeating_characters(s):
    # 初始化一个字典来保存字符及其在字符串中的索引。
    char_index = {}

    # 初始化变量max_length为0，用于记录最长不重复子字符串的长度。
    max_length = 0

    # 初始化变量start为0，表示当前考虑的子字符串的开始位置，这个变量会随着我们的遍历逐步向后移动。
    start = 0

    # 遍历字符串中的每一个字符，获取字符及其索引
    for i, char in enumerate(s):
        # 如果当前字符已经在字典中（即之前出现过），并且这个字符的位置在当前考虑的子字符串内，我们就需要移动子字符串的开始位置
        if char in char_index and start <= char_index[char]:
            start = char_index[char] + 1  # 移动开始位置到重复字符的下一位置

        # 如果当前字符是首次出现，或者出现的位置不在当前考虑的子字符串内，我们就可以继续考虑当前的子字符串。
        else:
            # 检查当前子字符串的长度是否大于最大长度，如果是，我们就更新最大长度。
            max_length = max(max_length, i - start + 1)

        # 不论当前字符是否重复，我们都需要更新字典中当前字符的位置。
        char_index[char] = i

    # 遍历完所有的字符后，返回最大长度。
    return max_length