2. 程式人生 > >[LeetCode] Minimum Cost to Merge Stones 混合石子的最小花費

[LeetCode] Minimum Cost to Merge Stones 混合石子的最小花費

阿新 • • 發佈：2019-03-08
return urn ons left cos task star turn it is



There are?N?piles of stones arranged in a row.? The?i-th pile has?stones[i]?stones.

A?move?consists of merging?exactly?K?consecutive?piles into one pile, and the cost of this move is equal to the total number of stones in these?K?piles.

Find the minimum cost to merge all piles of stones into one pile.? If it is impossible, return?-1

.

Example 1:

Input: stones = [3,2,4,1], K = 2
Output: 20
Explanation:
We start with [3, 2, 4, 1].
We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].
We merge [4, 1] for a cost of 5, and we are left with [5, 5].
We merge [5, 5] for a cost of 10, and we are left with [10].
The total cost was 20, and this is the minimum possible.

Example 2:

Input: stones = [3,2,4,1], K = 3
Output: -1
Explanation: After any merge operation, there are 2 piles left, and we can‘t merge anymore. So the task is impossible.

Example 3:

Input: stones = [3,5,1,2,6], K = 3
Output: 25
Explanation:
We start with [3, 5, 1, 2, 6].
We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].

We merge [3, 8, 6] for a cost of 17, and we are left with [17].
The total cost was 25, and this is the minimum possible.

Note:

  • 1 <= stones.length <= 30
  • 2 <= K <= 30
  • 1 <= stones[i] <= 100



參考資料：

https://leetcode.com/problems/minimum-cost-to-merge-stones/

[LeetCode] Minimum Cost to Merge Stones 混合石子的最小花費

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