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.
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)].
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)].
...
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];
}
}