leetcode981

考虑线性的搜索会超时，所以用二叉搜索来解决，代码如下：

class TimeMap:
    def __init__(self):
        self.ST = dict()

    def set(self, key: 'str', value: 'str', timestamp: 'int') -> 'None':
        if key in self.ST.keys():
            D = self.ST[key]#dict
            D.update({timestamp:value})
            self.ST.update({key:D})
        else:
            D = dict()
            D.update({timestamp:value})
            self.ST.update({key:D})


    def get(self, key: 'str', timestamp: 'int') -> 'str':
        if key in self.ST.keys():
            D = self.ST[key]
            V = self.binSearch(timestamp,D)
            return V
        else:
            return ''

    def binSearch(self,target,D):
        times = list(D.keys())
        n = len(times)
        minval = times[0]
        maxval = times[-1]
        if target < minval:
            return ''

        if target ==  minval:
            return D[times[0]]

        if target >= maxval:
            return D[times[-1]]

        left = 0
        right =  n - 1
        while left < right:
            mid = (left + right) // 2
            if times[mid] == target:
                return D[times[mid]]
            elif times[mid]<target:
                left = mid + 1
            elif times[mid]>target:
                right = mid - 1
        if left == 0:
            return D[times[left]]
        else:
            return D[times[left-1]]

但是这种写法会超时，这应该是代码质量问题，目前没明白是啥原因。

哪位博友知道我的代码的问题，欢迎告知。

参考了一下别人的方案，看到一个线性搜索的解决方案，却可以通过。

class TimeMap:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.dic = {}

    def set(self, key: 'str', value: 'str', timestamp: 'int') -> 'None':
        if key in self.dic:
            self.dic[key].append({'v': value, 't': timestamp})
        else:
            self.dic[key] = [{'v': value, 't': timestamp}]

    def get(self, key: 'str', timestamp: 'int') -> 'str':
        if key in self.dic:
            for kv in reversed(self.dic[key]):
                if timestamp >= kv['t']:
                    return kv['v']
            return ""
        else:
            return ""