p341

LeetCode p341 Flatten Nested List Iterator 题解

1.题目:

Given a nested list of integers, implement an iterator to flatten it.

Each element is either an integer, or a list – whose elements may also be integers or other lists.

Example 1:
Given the list [[1,1],2,[1,1]],

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,1,2,1,1].

Example 2:
Given the list [1,[4,[6]]],

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6].

题意:

实现一个数据结构

2.解题思路:

一层层脱壳,见代码。

3.代码


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/**
* // This is the interface that allows for creating nested lists.
* // You should not implement it, or speculate about its implementation
* public interface NestedInteger {
*
* // @return true if this NestedInteger holds a single integer, rather than a nested list.
* public boolean isInteger();
*
* // @return the single integer that this NestedInteger holds, if it holds a single integer
* // Return null if this NestedInteger holds a nested list
* public Integer getInteger();
*
* // @return the nested list that this NestedInteger holds, if it holds a nested list
* // Return null if this NestedInteger holds a single integer
* public List<NestedInteger> getList();
* }
*/
public class NestedIterator implements Iterator<Integer> {

private Stack<NestedInteger> stack = new Stack<NestedInteger>();


public NestedIterator(List<NestedInteger> nestedList) {

for (int i = nestedList.size() - 1; i > -1; i--) {
stack.push(nestedList.get(i));
}

}

@Override
public Integer next() {

return stack.pop().getInteger();

}

@Override
public boolean hasNext() {

while ( !stack.empty() && !stack.peek().isInteger()) {
NestedInteger nestedInteger = stack.pop();
List<NestedInteger> list = nestedInteger.getList();
for (int i = list.size() - 1; i > -1; i--) {
stack.push(list.get(i));
}
}
return !stack.empty();

}


}

/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i = new NestedIterator(nestedList);
* while (i.hasNext()) v[f()] = i.next();
*/

4.一些总结: