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.
class Solution {
public boolean stoneGame(int[] piles) {
int alex = 0, lee = 0;
int left = 0, right = piles.length - 1;
while(left < right) {
int alexcur = Math.max(piles[left], piles[right]);
alex += alexcur;
if(piles[left] == alexcur) {
lee += piles[right];
}
else lee += piles[left];
left++;
right--;
}
System.out.println(alex);
return alex > lee;
}
}1. 啊？？？后来觉得这么想错了，是每个人从首尾选一个，alex先选，但为什么还ac了？？？
return true;
2. 牛逼之直接return true，为啥？
我们知道一共有odd number总数的石头，分为even堆，那我们要是分一半，肯定会出现sum(odd index) 》 sum（even index）或者相反，所以我们每次让alex选odd or even index的不就行了？????，直接不演了
class Solution {
public boolean stoneGame(int[] p) {
int n = p.length;
int[][] dp = new int[n][n];
for (int i = 0; i < n; i++) dp[i][i] = p[i];//因为dp【i】【j】描述的是alex从i到j中比lee大多少,所以i==j时默认取了最大的（就是p【i】）
for (int d = 1; d < n; d++)
for (int i = 0; i < n - d; i++)
dp[i][i + d] = Math.max(p[i] - dp[i + 1][i + d], p[i + d] - dp[i][i + d - 1]);
return dp[0][n - 1] > 0;
}
}
3. dp，具体看lee大佬
