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题目
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:
2 <= piles.length <= 500
piles.length is even.
1 <= piles[i] <= 500
sum(piles) is odd.
题目地址
讲解
动态规划，好久没做动态规划了。三要素：最优子结构、无后效性、子问题重叠。核心思想是记录子问题的解（空间换时间）。具体做法有 自底向上（迭代，规模由小到大），自顶向下（递归，规模由大到小）。
Java 代码
自顶向下：
递归的比较好理解一点，状态转移方程：f(n) = {拿左边 +f(n-1) 左，拿右边 +f(n-1) 右 }
class Solution {
int beginIndex;
int endIndex;
int[][] dp;
public boolean stoneGame(int[] piles) {
beginIndex = 0;
endIndex = piles.length-1;
dp = new int[piles.length][piles.length];
for(int i=0;i<dp.length;i++){
for(int j=0;j<dp[i].length;j++){
dp[i][j]=-1;
}
}
int sum=0;
for(int x:piles){
sum += x;
}
return Math.max(piles[beginIndex]+stoneGame(piles, beginIndex+1, endIndex), piles[endIndex]+stoneGame(piles, beginIndex, endIndex-1))>sum/2?true:false;
}
private int stoneGame(int[] piles, int begin, int end){
if(begin>=end){
return 0;
}
if(dp[begin][end]!=-1){
return dp[begin][end];
}else{
dp[begin][end] = Math.max(piles[begin]+stoneGame(piles, begin+1, end), piles[end]+stoneGame(piles, begin, end-1));
return dp[begin][end];
}
}
}
自底向上：
把另一个人拿的数当做减。
求 piles[i] 到 piles[j] 的最优解，并存起来。
class Solution {
public boolean stoneGame(int[] piles) {
int[][] dp = new int[piles.length][piles.length];
for(int i=0;i<piles.length;i++){
for(int j=i+1;j<piles.length;j++){
int parity= (j-i)%2;
if(parity==1){
dp[i][j] = Math.max(piles[i]+dp[i+1][j], piles[j]+dp[i][j-1]);
}else{
dp[i][j] = Math.max(-piles[i]+dp[i+1][j], -piles[j]+dp[i][j-1]);
}
}
}
return dp[0][piles.length-1]>0;
}
}
