[변지협] Day19

leetcode3/변지협/v3/284. Peeking Iterator.py

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+'''
+1. 아이디어:
+그냥 문제에 주어진대로 구현하면 됨.
+2. 시간복잡도:
+o(n)
+3. 알고리즘:
+'''
+
+# Below is the interface for Iterator, which is already defined for you.
+#
+# class Iterator:
+#     def __init__(self, nums):
+#         """
+#         Initializes an iterator object to the beginning of a list.
+#         :type nums: List[int]
+#         """
+#
+#     def hasNext(self):
+#         """
+#         Returns true if the iteration has more elements.
+#         :rtype: bool
+#         """
+#
+#     def next(self):
+#         """
+#         Returns the next element in the iteration.
+#         :rtype: int
+#         """
+
+class PeekingIterator:
+    it = []
+    i = 0
+
+    def __init__(self, iterator):
+        """
+        Initialize your data structure here.
+        :type iterator: Iterator
+        """
+        tmp = []
+        while iterator.hasNext():
+            tmp.append(iterator.next())
+        self.it = tmp
+
+
+    def peek(self):
+        """
+        Returns the next element in the iteration without advancing the iterator.
+        :rtype: int
+        """
+        return self.it[self.i]
+
+    def next(self):
+        """
+        :rtype: int
+        """
+        tmp = self.it[self.i]
+        self.i +=1
+        return tmp
+
+
+    def hasNext(self):
+        """
+        :rtype: bool
+        """
+        try:
+            tmp = self.it[self.i]
+            return True
+        except:
+            return False
+
+# Your PeekingIterator object will be instantiated and called as such:
+# iter = PeekingIterator(Iterator(nums))
+# while iter.hasNext():
+#     val = iter.peek()   # Get the next element but not advance the iterator.
+#     iter.next()         # Should return the same value as [val].