写一个 resolve

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补全 resolve(merged, take)：把冲突块换成你要的那一半，并删掉三行标记。take 取 "ours" 或 "theirs"。

输出取对方版之后的行数。

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← 解决之后文件多少行 ⚠️ 检查有没有漏删标记 →

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def classify(b, o, t):
    if o == t:
        return "same"
    if o == b:
        return "theirs"
    if t == b:
        return "ours"
    return "conflict"

def merge3(base, ours, theirs):
    out = []
    for b, o, t in zip(base, ours, theirs):
        k = classify(b, o, t)
        if k == "conflict":
            out.append("<<<<<<< ours")
            out.append(o)
            out.append("=======")
            out.append(t)
            out.append(">>>>>>> theirs")
        elif k == "theirs":
            out.append(t)
        else:
            out.append(o)
    return out

def resolve(merged, take):
    out = []
    i = 0
    while i < len(merged):
        # TODO: 遇到 <<<<<<< 就找出 ======= 和 >>>>>>> 的位置
        # TODO: 按 take 取上半段或下半段，三行标记都不要
        # TODO: 不是冲突块的行照原样放进去
        i += 1
    return out

base = ['HOST = "localhost"', 'PORT = 8000', 'DEBUG = False',
        'TIMEOUT = 30', 'RETRY = 1', 'LOG = "app.log"']
ours = ['HOST = "localhost"', 'PORT = 8080', 'DEBUG = True',
        'TIMEOUT = 60', 'RETRY = 1', 'LOG = "main.log"']
theirs = ['HOST = "0.0.0.0"', 'PORT = 9000', 'DEBUG = True',
          'TIMEOUT = 30', 'RETRY = 3', 'LOG = "sys.log"']
m = merge3(base, ours, theirs)
print(len(resolve(m, "theirs")))

本次输入：

输出：

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