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:
Given the list [[1,1],2,[1,1]], return 10. (four 1's at depth 2, one 2 at depth 1)

Example 2:
Given the list [1,[4,[6]]], return 27. (one 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 42 + 63 = 27)

BFS

Time Complexity
O(N)
Space Complexity
O(N)

思路

It is a BFS problem. We can image this nested list as a tree.
For example, [[1,1],2,[1,1]]

So it is a easy BFS traversal. :)

代码

public int depthSum(List<NestedInteger> nestedList) {
    if(nestedList == null || nestedList.size() == 0) return 0;
    Queue<NestedInteger> queue = new LinkedList<NestedInteger>();
    for(NestedInteger nest : nestedList) queue.offer(nest);
    int level = 1, sum = 0;
    while(!queue.isEmpty()){
        int size = queue.size();
        int levelSum = 0;
        for(int i = 0; i < size; i++){
            NestedInteger cur = queue.poll();
            if(cur.isInteger()){
                levelSum += cur.getInteger() * level;
            }else{
                for(NestedInteger n : cur.getList()){
                    queue.offer(n);
                }
            }
        }
        sum += levelSum;
        level++;
    }
    return sum;
}

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; 13 + 42 + 6*1 = 17)

BFS

Time Complexity
O(N)
Space Complexity
O(N)

思路

Just change the previous code a little bit. It looks like we need to know the depth of the level first, but we just need to keep a global level sum.
Every time add one level, just add one more level sum to the total sum.

代码

public int depthSumInverse(List<NestedInteger> nestedList) {
    if(nestedList == null || nestedList.size() == 0) return 0;
    Queue<NestedInteger> queue = new LinkedList<NestedInteger>();
    for(NestedInteger n : nestedList){
        queue.offer(n);
    }
    int sum = 0, levelSum = 0;
    while(!queue.isEmpty()){
        int size = queue.size();
        for(int i = 0; i < size; i++){
            NestedInteger cur = queue.poll();
            if(cur.isInteger()){
                levelSum += cur.getInteger();
            }else{
                for(NestedInteger n : cur.getList()){
                    queue.offer(n);
                }
            }
        }
        sum += levelSum;
    }
    return sum;
}

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