java Java迭代器在不增加的情况下获得下一个
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Java iterator get next without incrementing
提问by Dzung Nguyen
I am writing the following loop in Java, for each loop I want to access the current and the next element of linked list r:
我正在用 Java 编写以下循环，对于每个循环，我想访问链表 r 的当前元素和下一个元素：
List<T> r = new LinkedList();
for (int i=0; i < r.size() - 1; i++) {
T current = r.get(i);
T next = r.get(i+1);
}
This might be wasting as everytime I call get(i), it start from the beginning so the run time order of the code is O(n^2). How do I achieve the same using Iterator (this time it will be O(n))? This is my first attempt:
这可能是浪费，因为每次我调用 get(i) 时，它都会从头开始，因此代码的运行时顺序是 O(n^2)。我如何使用迭代器实现相同的目标（这次是 O(n)）？这是我的第一次尝试：
while(it.hasNext()) {
T current = it;
T next = it.next();
}
回答by rgettman
Maintain a variable previousthat is equal to the previous loop's currentvalue.
维护一个previous等于前一个循环current值的变量。
T previous = null;
// If it makes sense to skip the first "null, first element" pair...
if (it.hasNext())
{
previous = it.next();
}
while (it.hasNext())
{
T current = it.next();
// Process previous and current here.
// End of loop, after processing. Maintain previous reference.
previous = current;
}
This will be O(n) because you are using the Iteratorover your entire linked list.
这将是 O(n)，因为您正在使用Iterator整个链表。
回答by RealSkeptic
In each iteration, you should keep around a variable that will be the "current" and one that will be the "next". And you start processing your information starting from the seconditeration, when you already have a currentsaved from the previous round.
在每次迭代中，您应该保留一个变量，即“当前”变量和“下一个”变量。然后您从第二次迭代开始处理您的信息，此时您已经current保存了上一轮的信息。
T current = null;
T next = null;
Iterator<T> iter = r.iterator();
while ( iter.hasNext() ) {
next = iter.next();
if ( current != null ) { // Passed the first iteration
// Now you can use current and next, they both have values.
}
current = next; // Save what was the "next" as the next "current".
}
It's best to make sure that the list itself doesn't have null values. If it does, and it's a valid value, then you should have a boolean flag instead of just checking whether current != null.
最好确保列表本身没有空值。如果是这样，并且它是一个有效值，那么您应该有一个布尔标志，而不仅仅是检查current != null.
回答by FredK
T current = r.get(0);
for ( int i=0; i < r.size()-1; i++ ) {
T next = r.get(i+1);
// do stuiff here
current = next;
}
