341. Flatten Nested List Iterator展开多层数组

［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］：

［思惟问题］：

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： 正常状况：特大：特小：程序里处理到的特殊状况：异常状况（不合法不合理的输入）：

［画图］：

［一刷］：

1. .next创建在.hasnext的基础之上，因此hasnext须要放在while循环中，作完为止
2. curr若是是个list，就必须用专有方法

   curr.getList()

   先取出，再作后续操做

［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，为何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。因此用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));
    }

［其余解法］：

［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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