My code:
/**
* // 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 unweighted = 0;
int weighted = 0;
List<NestedInteger> next = new ArrayList<NestedInteger>();
while (nestedList.size() != 0) {
for (NestedInteger temp : nestedList) {
if (temp.isInteger()) {
unweighted += temp.getInteger();
}
else {
next.addAll(temp.getList());
}
}
weighted += unweighted;
nestedList = next;
next = new ArrayList<NestedInteger>();
}
return weighted;
}
}
这道题目并没有做出来。。。
觉得会很复杂。看答案,没想到思路如此简洁简单。
每次总是这样,自己觉得很复杂的问题,他们用很简洁的思路就做了出来,那些我考虑的恶心的东西,他们根本不考虑。
reference:
https://discuss.leetcode.com/topic/49041/no-depth-variable-no-multiplication
关键在于,
即使一个数在他的那一枝是叶子,不代表他可以 乘以 1
因为如果还有更深的结点,那他的层数就大于 1
所以可以用累加的方式。
比如最开始一层是 5,我们无法确定它的深度,
那就每一层都加上他。这样就一定能保证它的深度正确。
Anyway, Good luck, Richardo! — 09/10/2016