1000.Minimum-Cost-to-Merge-Stones

1000. Minimum Cost to Merge Stones

题目地址

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

题目描述

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

代码

Approach #1

class Solution {
    public int mergeStones(int[] stones, int K) {
    int n = stones.length;
    if ((n - 1) % (K - 1) != 0)    return -1;

    int[] sum = new int[n + 1];
    for (int i = 0; i < n; i++) {
      sum[i + 1] = sum[i] + stones[i];
    }

    int[][] dp = new int[n][n];
    for (int len = K; len <= n; len++) {
      for (int i = 0; i + len <= n; i++) {
        int j = i + len - 1;
        dp[i][j] = Integer.MAX_VALUE;
        for (int mid = i; mid < j; mid += K - 1) {
          dp[i][j] = Math.min(dp[i][j], dp[i][mid] + dp[mid + 1][j]);
        }
        if ((j - i) % (K - 1) == 0) {
          dp[i][j] += sum[j + 1] - sum[i];
        }
      }
    }

    return dp[0][n - 1];
  }
}