Nested List Weight Sum
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.
Example 1:
Input: [[1,1],2,[1,1]]
Output: 10
Explanation: Four 1's at depth 2, one 2 at depth 1.
Example 2:
Input: [1,[4,[6]]]
Output: 27
Explanation: One 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 4*2 + 6*3 = 27.
这道题定义了一种嵌套链表的结构,链表可以无限往里嵌套,规定每嵌套一层,深度加1,让我们求权重之和,就是每个数字乘以其权重,再求总和。那么我们考虑,由于嵌套层数可以很大,所以我们用深度优先搜索DFS会很简单,每次遇到嵌套的,递归调用函数,一层一层往里算就可以了,我最先想的方法是遍历给的嵌套链表的数组,对于每个嵌套链表的对象,调用getSum函数,并赋深度值1,累加起来返回。在getSum函数中,首先判断其是否为整数,如果是,则返回当前深度乘以整数,如果不是,那么我们再遍历嵌套数组,对每个嵌套链表再调用递归函数,将返回值累加起来返回即可,
/**
* // 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();
* }
*/
class Solution {
public int depthSum(List<NestedInteger> nestedList) {
int res = 0;
for(NestedInteger iterm : nestedList){
res += getSum(iterm,1);
}
return res;
}
public int getSum(NestedInteger ni, int level){
int res = 0;
if(ni.isInteger())
return ni.getInteger()*level;
for(NestedInteger item : ni.getList()){
res += getSum(item,level+1);
}
return res;
}
}
优化到时间复杂度 o(n),直接在while循环里就进行操作
/**
* // 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();
* }
*/
class Solution {
public int depthSum(List<NestedInteger> nestedList) {
if(nestedList == null || nestedList.size() == 0){
return 0;
}
//first round offer to queue;
Queue<NestedInteger> queue = new LinkedList<>();
for(NestedInteger ni : nestedList){
queue.offer(ni);
}
int level = 1, sum = 0;
while(!queue.isEmpty()){
int size = queue.size();
int levelSum = 0;
for(int i = 0; i < size;i++){
NestedInteger ni = queue.poll();
if(ni.isInteger()){
levelSum += ni.getInteger()*level;
}else{
for(NestedInteger item : ni.getList()){
queue.offer(item);
}
}
}
sum += levelSum;
level++;
}
return sum;
}
}
犯的错:
levelSum 写到了while循环外面,导致每次循环时 levelSum没有归零
while循环内判断item不是数字的时候,直接把item加进了queue,导致了死循环,应该把item的list拿出来,遍历往queue里加
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