364. Nested List Weight Sum II

Given a nested list of integers, return the sum of all integers in the list weighted by their depth.

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

Different from the previous question where weight is increasing from root to leaf, now the weight
is defined from bottom up. i.e., the leaf level integers have weight 1, and the root level integers have the largest weight.

Example 1:

Given the list 

[[1,1],2,[1,1]]

, return 8.
(four 1's at depth 1, one 2 at depth 2)

Example 2:

Given the list 

[1,[4,[6]]]

, return 17.
(one 1 at depth 3, one 4 at depth 2, and one 6 at depth 1; 1*3 + 4*2 + 6*1 = 17)
用迭代的方法解题。新建一个list存放下一层的list，因为权重是从下往上算，以n层的嵌套，第一层的一个integer为例，在结果中被加了n次，第二层被加了n-1次，所以新建一个currsum，里面有所有的integer，遇到一个integer就加进来，在这一层循环的最后，加上这个currsum，等于加了一次，下一层，这些数又会被加一次，到最底下时，第一层的数就被加了n次，最后返回sum。代码如下：
/**
* // This is the interface that allows for creating nested lists.
* // You should not implement it, or speculate about its implementation
* public interface NestedInteger {
* // Constructor initializes an empty nested list.
* public NestedInteger();
*
* // Constructor initializes a single integer.
* public NestedInteger(int value);
*
* // @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();
*
* // Set this NestedInteger to hold a single integer.
* public void setInteger(int value);
*
* // Set this NestedInteger to hold a nested list and adds a nested integer to it.
* public void add(NestedInteger ni);
*
* // @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 Solution {
public int depthSumInverse(List<NestedInteger> nestedList) {
int weight = 0, unweight = 0;
while (!nestedList.isEmpty()) {
List<NestedInteger> nextLevel= new ArrayList<NestedInteger>();
for (NestedInteger nested: nestedList) {
if (nested.isInteger()) {
unweight += nested.getInteger();
} else {
nextLevel.addAll(nested.getList());
}
}
weight += unweight;
nestedList = nextLevel;
}
return weight;
}
}也可以用queue提高速度，不用每次都改变nestedlist。代码如下：
public class Solution {
public int depthSumInverse(List<NestedInteger> nestedList) {
int sum = 0;
int currSum = 0;
int size = nestedList.size();
Queue<NestedInteger> list = new LinkedList<>(nestedList);
while (size > 0) {
int nextSize = 0;
for (int i = 0; i < size; i++) {
NestedInteger n = list.poll();
if (n.isInteger()) {
currSum += n.getInteger();
} else {
list.addAll(n.getList());
nextSize += n.getList().size();
}
}
sum += currSum;
size = nextSize;
}
return sum;
}
}