[抄题]:

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].

[暴力解法]:

时间分析:

空间分析:

[优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

[思维问题]:

不知道.next 和 .hasnext有啥区别:取出来、只是看看有没有

[一句话思路]:

只有stack才能一次取出来一层,getlist getinteger

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

  1. .next建立在.hasnext的基础之上,所以hasnext需要放在while循环中,做完为止
  2. curr如果是个list,就必须用专有方法
    curr.getList()

    先取出,再做后续操作

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

一次放一层

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[英文数据结构或算法,为什么不用别的数据结构或算法]:

list 对应的方法是.size() .get

只有stack才能一次取出来一层,数组不能直接取出来一层。所以用stack。

stack有

.getInteger()
.getList()

方法

[算法思想:递归/分治/贪心]:

[关键模板化代码]:

public NestedIterator(List<NestedInteger> nestedList) {
//put into stack from back
for (int i = nestedList.size() - 1; i >= 0; i--) stack.push(nestedList.get(i));
}

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

[代码风格] :

/**
* // 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> {
//ini:stack
Stack<NestedInteger> stack = new Stack<>(); public NestedIterator(List<NestedInteger> nestedList) {
//put into stack from back
for (int i = nestedList.size() - 1; i >= 0; i--) stack.push(nestedList.get(i));
} @Override
public Integer next() {
//pop
return stack.pop().getInteger();
} @Override
public boolean hasNext() {
while (!stack.isEmpty()) {
//isInteger or put into stack from back
NestedInteger curr = stack.peek();
if (curr.isInteger()) return true; stack.pop();
for (int i = curr.getList().size() - 1; i >= 0; i--) {
stack.push(curr.getList().get(i));
} }
return false;
}
} /**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i = new NestedIterator(nestedList);
* while (i.hasNext()) v[f()] = i.next();
*/

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