# 877. Stone Game 石子游戏

@TOC

## # 题目描述

Alex and Lee play a game with piles of stones. There are an even 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. The total number of stones is odd, so there are no ties.

Alex and Lee take turns, with Alex starting first. Each turn, a player takes the entire pile of stones from either the beginning or the end of the row. This continues until there are no more piles left, at which point the person with the most stones wins.

Assuming Alex and Lee play optimally, return True if and only if Alex wins the game.

Example 1:

``````Input: [5,3,4,5]
Output: true
Explanation:
Alex starts first, and can only take the first 5 or the last 5.
Say he takes the first 5, so that the row becomes [3, 4, 5].
If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points.
If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alex, so we return true.
``````

Note:

1. 2 <= piles.length <= 500
2. piles.length is even.
3. 1 <= piles[i] <= 500
4. sum(piles) is odd.

## # 解题方法

### # 数学

``````class Solution:
def stoneGame(self, piles):
return True
``````

### # 双函数

alex是先选的，所以调用f函数判断他能否赢。

``````class Solution(object):

def stoneGame(self, piles):
"""
:type piles: List[int]
:rtype: bool
"""
if not piles:
return False
self.F = [[0 for i in range(len(piles))] for j in range(len(piles))]
self.S = [[0 for i in range(len(piles))] for j in range(len(piles))]
_sum = sum(piles)
alex = self.f(piles, 0, len(piles) - 1)
return alex > _sum / 2

def f(self, piles, i, j):
"""
先选
"""
if i == j:
return piles[i]
if self.F[i][j] != 0:
return self.F[i][j]
curr = max(piles[i] + self.s(piles, i + 1, j), piles[j] + self.s(piles, i, j - 1))
self.F[i][j] = curr
return curr

def s(self, piles, i, j):
"""
后选
"""
if i == j:
return 0
if self.S[i][j] != 0:
return self.S[i][j]
curr = min(self.f(piles, i + 1, j), self.f(piles, i, j - 1))
self.S[i][j] = curr
return curr
``````

``````class Solution(object):

def stoneGame(self, piles):
"""
:type piles: List[int]
:rtype: bool
"""
self.f_map, self.s_map = dict(), dict()
_sum = sum(piles)
alex = self.f(piles, 0, len(piles)-1)
print(alex, _sum)
return alex > _sum / 2.0

def f(self, piles, start, end):
if start == end:
return piles[start]
if (start, end) not in self.f_map:
f_val = max(piles[start] + self.s(piles, start+1, end), piles[end] + self.s(piles, start, end-1))
self.f_map[(start, end)] = f_val
return self.f_map[(start, end)]

def s(self, piles, start, end):
if start == end:
return 0
if (start, end) not in self.s_map:
s_val = min(self.f(piles, start+1, end), self.f(piles, start, end-1))
self.s_map[(start, end)] = s_val
return self.s_map[(start, end)]
``````

### # 单函数 + 记忆化递归

``````class Solution {
public:
bool stoneGame(vector<int>& piles) {
const int N = piles.size();
m_ = vector<vector<int>>(N, vector<int>(N, INT_MIN));
return score(piles, 0, N - 1) > 0;
}
private:
vector<vector<int>> m_;
//Alex比Lee多的分数
int score(vector<int>& piles, int l, int r) {
if (l == r) return piles[l];
if (m_[l][r] == INT_MIN) {
m_[l][r] = max(piles[l] - score(piles, l + 1, r),
piles[r] - score(piles, l, r - 1));
}
return m_[l][r];
}
};
``````

### # 动态规划

``````class Solution {
public:
bool stoneGame(vector<int>& piles) {
const int N = piles.size();
// dp[i] := max(your_stones - op_stones) for piles[i] to piles[i + l - 1]
vector<vector<int>> dp(N, vector<int>(N, INT_MIN));
for (int i = 0; i < N; i++)
dp[i][i] = piles[i];
for (int l = 2; l <= N; l++) {
for (int i = 0; i < N - l + 1; i++) {
int j = i + l - 1;
dp[i][j] = max(piles[i] - dp[i + 1][j], piles[j] - dp[i][j - 1]);
}
}
return dp[0][N - 1] > 0;
}
};
``````

https://leetcode.com/problems/stone-game/discuss/154610/C++JavaPython-DP-or-Just-return-true

## # 日期

2018 年 9 月 4 日 —— 迎接明媚的阳光！ 2018 年 12 月 4 日 —— 周二啦！