亚洲精品不卡AV在线播放|国产性爱无码高清视频|国产成人 ,97人人色,国产免费一区二区三区,日本一区二区在线

To solve this problem, we need to split an input string into pairs of two characters each. If the string has an odd length, the last pair should consist of the remaining character followed by an underscore.

Approach

1. Initialize an empty list to store the resulting pairs.
2. Iterate through the string in steps of 2: This allows us to process each character and its next one (if it exists) as a pair.
3. Check for valid pairs: For each position i:
   • If there is a next character (i.e., i+1 < len(s)), form a pair with the current and next character.
   • If there is no next character (end of string with odd length), form a pair with the current character and an underscore.
4. Return the list of pairs.

Solution Code

def solution(s):
    res = []
    for i in range(0, len(s), 2):
        if i + 1 < len(s):
            res.append(s[i] + s[i+1])
        else:
            res.append(s[i] + '_')
    return res

Explanation

• Loop through the string: The loop runs from index 0 to the length of the string, incrementing by 2 each time. This ensures we process two characters at a time.
• Form pairs: For each index i, we check if the next index (i+1) is within bounds. If yes, we add the pair s[i] + s[i+1] to the result list. If not, we add s[i] + '_' to handle the odd-length case.
• Edge cases: Handles empty strings (returns an empty list), even-length strings (returns full pairs), and odd-length strings (adds underscore to the last character).

This approach efficiently processes the string in linear time, O(n), where n is the length of the input string, making it optimal for all valid input sizes. The space complexity is O(n) to store the resulting pairs.

Answer: The code as provided is the correct solution.

def solution(s):
    res = []
    for i in range(0, len(s), 2):
        if i+1 < len(s):
            res.append(s[i]+s[i+1])
        else:
            res.append(s[i] + '_')
    return res

（免責聲明：本文為本網(wǎng)站出于傳播商業(yè)信息之目的進行轉(zhuǎn)載發(fā)布，不代表本網(wǎng)站的觀點及立場。本文所涉文、圖、音視頻等資料的一切權(quán)利和法律責任歸材料提供方所有和承擔。本網(wǎng)站對此資訊文字、圖片等所有信息的真實性不作任何保證或承諾，亦不構(gòu)成任何購買、投資等建議，據(jù)此操作者風險自擔。） 本文為轉(zhuǎn)載內(nèi)容，授權(quán)事宜請聯(lián)系原著作權(quán)人，如有侵權(quán)，請聯(lián)系本網(wǎng)進行刪除。