原题链接在这里：
题目：
Given an array arr of positive integers, consider all binary trees such that:
- Each node has either 0 or 2 children;
- The values of
arrcorrespond to the values of each leaf in an in-order traversal of the tree. (Recall that a node is a leaf if and only if it has 0 children.) - The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree respectively.
Among all possible binary trees considered, return the smallest possible sum of the values of each non-leaf node. It is guaranteed this sum fits into a 32-bit integer.
Example 1:
Input: arr = [6,2,4]Output: 32Explanation:There are two possible trees. The first has non-leaf node sum 36, and the second has non-leaf node sum 32. 24 24 / \ / \ 12 4 6 8 / \ / \6 2 2 4
Constraints:
2 <= arr.length <= 401 <= arr[i] <= 15- It is guaranteed that the answer fits into a 32-bit signed integer (ie. it is less than
2^31).
题解：
From the example, see that it is better to remove smaller element first.
Remove small element i and the cost is arr[i] * Math.min(arr[i-1], arr[i+1]). minimum cost happens between smaller values of i-1 and i+1.
Remove until there is only one element and sum of cost is the answer.
Use stack to maintain decreasing order, when there is bigger value num, then pop small value arr[i] and acculate the cost arr[i] * Math.min(num, stk.peek()).
Time Complexity: O(n). n = arr.length.
Space: O(n).
AC Java:
1 class Solution { 2 public int mctFromLeafValues(int[] arr) { 3 if(arr == null || arr.length < 2){ 4 return 0; 5 } 6 7 int res = 0; 8 Stack stk = new Stack<>(); 9 stk.push(Integer.MAX_VALUE);10 for(int num : arr){11 while(stk.peek() <= num){12 int mid = stk.pop();13 res += mid*Math.min(stk.peek(), num);14 }15 16 stk.push(num);17 }18 19 while(stk.size() > 2){20 res += stk.pop()*stk.peek();21 }22 23 return res;24 }25 } 跟上.