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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?
Given array A = [3,2,2], return true.
Given array A = [1,2,4], return true.
Given array A = [1,20,4], return false.
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 };
[LintCode] Coins in a Line III
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原文地址：http://www.cnblogs.com/easonliu/p/4741058.html
