标签：time 很多 amount lang length param cto hal exp
There are n coins in a line, and value of i-th coin is values[i].
Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Could you please decide the first player will win or lose?
Example 1:
Input: [3, 2, 2]
Output: true
Explanation: The first player takes 3 at first. Then they both take 2.Example 2:
Input: [1, 20, 4]
Output: false
Explanation: The second player will take 20 whether the first player take 1 or 4.
O(1) memory and O(n) time when n is even.
思路：
区间动态规划问题, 具体定义状态的方式有很多种, 但是大同小异, 时空复杂度都相似. 这里只介绍 version 1 的具体实现.
设定 dp[i][j] 表示当前剩余硬币的区间为 [i, j] 时, 此时该拿硬币的人能获取的最大值.
注意, dp[i][j] 并没有包含角色信息, dp[0][values.length - 1] 表示的是先手的人能获得的最大值, 而 dp[1][values.length -1] 表示的则是后手的人能获得的最大值. 需要这样做是因为: 两个人都会采用最优策略.
当前的人的决策就是拿左边还是拿右边, 而下一个人仍然会最优决策, 所以应该是最小值中取最大值:
dp[i][j] = max( // 取max表示当前的人选用最优策略
min(dp[i + 2][j], dp[i + 1][j - 1]) + values[i], // 取min表示下一个人选用最优策略
min(dp[i][j - 2], dp[i + 1][j - 1]) + values[j]
)
几个边界:
i > j : dp[i][j] = 0
i == j : dp[i][j] = values[i]
i + 1 == j : dp[i][j] = max(values[i], values[j])public class Solution {
/**
* @param values: a vector of integers
* @return: a boolean which equals to true if the first player will win
*/
public boolean firstWillWin(int[] values) {
int n = values.length;
if (n == 0) {
return true;
}
int[][] f = new int[n][n];
int i, j, len;
// len 1
for (i = 0; i < n; ++i) {
f[i][i] = values[i];
}
for (len = 2; len <= n; ++len) {
for (i = 0; i <= n - len; ++i) {
j = i + len - 1;
f[i][j] = Math.max(values[i] - f[i + 1][j], values[j] - f[i][j - 1]);
}
}
return f[0][n - 1] >= 0;
}
}
标签：time 很多 amount lang length param cto hal exp
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078331.html
