zoukankan      html  css  js  c++  java
  • [LintCode] Coins in a Line III

    Coins in a Line III

    There are n coins in a line. 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

    Given array A = [3,2,2], return true.

    Given array A = [1,2,4], return true.

    Given array A = [1,20,4], return false.

    Challenge

    Follow Up Question:

    If n is even. Is there any hacky algorithm that can decide whether first player will win or lose in O(1) memory and O(n) time?

    备忘录,dp[left][right]表示从left到right所能取到的最大值,因为双方都取最优策略,所以取完一个后,对手有两种选择,我们要加让对手得到更多value的那种方案,也就是自己得到更少value的方案。

     1 class Solution {
     2 public:
     3     /**
     4      * @param values: a vector of integers
     5      * @return: a boolean which equals to true if the first player will win
     6      */
     7     bool firstWillWin(vector<int> &values) {
     8         // write your code here
     9         int n = values.size();
    10         vector<vector<int>> dp(n + 1, vector<int>(n + 1, -1));
    11         int sum = 0;
    12         for (auto v : values) sum += v;
    13         return sum < 2 * dfs(dp, values, 0, n - 1);
    14     }
    15     int dfs(vector<vector<int>> &dp, vector<int> &values, int left, int right) {
    16         if (dp[left][right] != -1) return dp[left][right];
    17         if (left == right) {
    18             dp[left][right] = values[left];
    19         } else if (left > right) {
    20             dp[left][right] = 0;
    21         } else {
    22             int take_left = min(dfs(dp, values, left + 2, right), dfs(dp, values, left + 1, right - 1)) + values[left];
    23             int take_right = min(dfs(dp, values, left, right - 2), dfs(dp, values, left + 1, right - 1)) + values[right];
    24             dp[left][right] = max(take_left, take_right);
    25         }
    26         return dp[left][right];
    27     }
    28 };
  • 相关阅读:
    springboot访问redis序列化问题
    观察者模式在源码中的使用
    抽象工厂模式在源码中的使用
    工厂方法模式在源码中的应用
    策略模式的应用
    springcloud实现限流
    Elasticsearch
    java8新特性(二)StreamApi
    thinkpaidE450 win10 进入bios
    restframe_work2
  • 原文地址:https://www.cnblogs.com/easonliu/p/4741058.html
Copyright © 2011-2022 走看看