给你一个嵌套的整型列表。请你设计一个迭代器,使其能够遍历这个整型列表中的所有整数。 列表中的每一项或者为一个整数,或者是另一个列表。其中列表的元素也可能是整数或是其他列表。 示例 1: 输入: [[1,1],2,[1,1]] 输出: [1,1,2,1,1] 解释: 通过重复调用 next 直到 hasNext 返回 false,next 返回的元素的顺序应该是: [1,1,2,1,1]。 示例 2: 输入: [1,[4,[6]]] 输出: [1,4,6] 解释: 通过重复调用 next 直到 hasNext 返回 false,next 返回的元素的顺序应该是: [1,4,6]。
题解:
/** * // 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> { List<Integer> list = new ArrayList<>(); int index = 0; public void add(List<NestedInteger> nestedList) { for (NestedInteger i : nestedList) { if (i.isInteger()) { list.add(i.getInteger()); } else { add(i.getList()); } } } public NestedIterator(List<NestedInteger> nestedList) { add(nestedList); } @Override public Integer next() { Integer i = list.get(index); index ++; return i; } @Override public boolean hasNext() { return index < list.size(); } } /** * Your NestedIterator object will be instantiated and called as such: * NestedIterator i = new NestedIterator(nestedList); * while (i.hasNext()) v[f()] = i.next(); */
分析: 这题最难的其实是理解题意。。。