1140.Stone-Game-II

1140. Stone Game II

题目地址

https://leetcode.com/problems/stone-game-ii/ https://www.acwing.com/solution/LeetCode/content/3172/

题目描述

Alex and Lee continue their games with piles of stones.  There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].  The objective of the game is to end with the most stones. 

Alex and Lee take turns, with Alex starting first.  Initially, M = 1.

On each player's turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M.  Then, we set M = max(M, X).

The game continues until all the stones have been taken.

Assuming Alex and Lee play optimally, return the maximum number of stones Alex can get.

Example 1:
Input: piles = [2,7,9,4,4]
Output: 10
Explanation:  If Alex takes one pile at the beginning, Lee takes two piles, then Alex takes 2 piles again. Alex can get 2 + 4 + 4 = 10 piles in total. If Alex takes two piles at the beginning, then Lee can take all three piles left. In this case, Alex get 2 + 7 = 9 piles in total. So we return 10 since it's larger. 

Constraints:
1 <= piles.length <= 100
1 <= piles[i] <= 10 ^ 4

代码

Approach #1 DFS with memorization

Time: O(1) && Space: O(1)

class Solution {
  int[] sums;
  int[][] dp;
  public int stoneGameII(int[] piles) {
        if (piles == null || piles.length == 0) {
      return 0;
    }
    int n = piles.length;
    dp = new int[n][n];

    sums = new int[n];
    sums[n-1] = piles[n-1];
    for (int i = n - 2; i >= 0; i--) {
      sums[i] = sums[i+1] + piles[i];
    }
    return helper(piles, 0, 1);
  }

  private int helper(int[] a, int i, int M) {
    if (i == a.length)        return 0;
    if (2*M >= a.length - i) {
      return sum[i];
    }
    if (dp[i][M] != 0)        return dp[i][M];
    int min = Integer.MAX_VALUE; // the min value the next player can get
    for (int x = 1; x <= 2 * M; x++) {
      min = Math.min(min, helper(a, i+x, Math.max(M, x));)
    }
    dp[i][M] = sums[i] - min; // max stones = all the left stones - the min stones next player can get
    return dp[i][M];
  }
}

Approach #2 DP

Let DP[i][m] be the maximal number of stones a player can get when i is the current position and the current M is m.

1. If the player takes only the first pile (piles[i], x = 1), the other player can get up to DP[i+1][max(m, 1)]. So the current player can get Si - DP[i+1][max(m, 1)].

2. If the player takes the first two piles (piles[i], piles[i+1], x = 2), the other player can get up to DP[i+2][max(m, 2)]. So the current player can get Si - DP[i+2][max(m, 2)].

3. ...

4. If the player takes the first 2m piles (piles[i], piles[i+1], ..., piles[i+2m], x = 2m), the other player can get up to DP[i+2m][max(m, 2m)]. So the current player can get Si - DP[i+2m][max(m, 2m)].

Time: O(n^3), 状态数O(n^2)，转移需要O(n)的时间 Space: O(n^2)

class Solution {
  public int stoneGameII(int[] piles) {
    int n = piles.length;
    for (int i  = n - 2; i >= 0; i--) {
      piles[i] += piles[i+1];
    }
    if (n <= 2) {
      return piles[0];
    }
    int[][] dp = new int[n][(n+1)/2 + 1];
    for (int i = n-1; i >= 0; i--) {
      int sum = piles[i];
      int m = (n-i+1) / 2;
      dp[i][m] = sum;
      while (--m >= 1) {
        dp[i][m] = 0;
        for (int x = 1; x <= m*2 && i+x < n; x++) {
          int max = Math.min((n-i-x+1)/2, Math.max(x, m));
          dp[i][m] = Math.max(dp[i][m], sum - dp[i+x][max]);
        }
      }
    }
    return dp[0][1];
  }
}