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:
Input: [[1,1],2,[1,1]]
Output: [1,1,2,1,1]
Explanation: By calling next repeatedly until hasNext returns false,
the order of elements returned by next should be: [1,1,2,1,1]
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Example 2:
Input: [1,[4,[6]]] Output: [1,4,6] Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be:[1,4,6]
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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 LinkedList<NestedInteger> stack; public NestedIterator(List<NestedInteger> nestedList) { stack = new LinkedList<>(); pushToStack(nestedList); } @Override public Integer next() { return hasNext() ? stack.pollFirst().getInteger() : null; } @Override public boolean hasNext() { while (!stack.isEmpty()) { NestedInteger cur = stack.peekFirst(); if (cur.isInteger()) { return true; } pushToStack(stack.pollFirst().getList()); } return false; } private void pushToStack(List<NestedInteger> nestedList) { // add in reverse order to make pop easy for (int i = nestedList.size() - 1; i >= 0; i--) { stack.offerFirst(nestedList.get(i)); } } } /** * Your NestedIterator object will be instantiated and called as such: * NestedIterator i = new NestedIterator(nestedList); * while (i.hasNext()) v[f()] = i.next(); */