leetcode_486. Predict the Winner

https://leetcode.com/problems/predict-the-winner/

题目描述：给定一个非负的积分数组，玩家1可以从数组两端任取一个积分，接着玩家2执行同样的操作，直至积分被取尽，总分大的获胜。两人都以最优决策进行游戏。对每个数组输出玩家1是否能获胜。

解法1：

使用递归，两者依次取数。

class Solution{
public:bool PredictTheWinner(vector<int>& nums){return dfs(nums, 0, nums.size()-1, 0, 0, 0);}bool dfs(vector<int>& nums, int st, int en, int p1, int p2, bool role){if(st == en){if(!role && p1+nums[st]>=p2)return true;else if(role && p1<p2+nums[st])return true;elsereturn false;}if(!role){bool c1 = dfs(nums, st+1,en,p1+nums[st],p2,!role);bool c2 = dfs(nums, st,en-1,p1+nums[en],p2,!role);if(c1 && c2)  //若从任意一端取数后，对手都胜，那么当前玩家必败return false;elsereturn true;}else{bool c1 = dfs(nums, st+1,en,p1,p2+nums[st],!role);bool c2 = dfs(nums, st,en-1,p1,p2+nums[en],!role);if(c1 && c2)return false;elsereturn true;}}
};

解法二：官方题解

对于两个玩家而言，玩家1的总分s1，玩家2的总分s2，dis=s1-s2，若玩家1胜，dis>=0，否则dis<0。依旧使用递归，双方依次取数，玩家1希望差值越大越好，玩家2希望差值越小越好。

int dfs(vector<int>& nums, int st, int en, bool role)函数返回值为：在nums[st,en]上由role取数后的总分差值dis。

class Solution{
public:bool PredictTheWinner(vector<int>& nums){return dfs(nums, 0, nums.size()-1, 0)>=0;}int dfs(vector<int>& nums, int st, int en, bool role){if(st == en){if(role == 0)return nums[st];elsereturn -nums[st];}if(!role)return max(nums[st] + dfs(nums, st+1, en, !role), nums[en]+dfs(nums, st, en-1, !role));elsereturn min(-nums[st] + dfs(nums, st+1, en, !role), -nums[en]+dfs(nums, st, en-1, !role));}
};

解法三：官方题解，动态规划

dp[st][en]：玩家1在nums[st,en]上取数过后的差值dis(dis=s1-s2)

在nums[st][en]上的dis: dp[st][en]取决于{num[st], dp[st+1][en]}和{num[en], dp[st][en-1]}，即仅取决于nums[st][en]子数组上的dp和num[st],num[en]。

class Solution{
public:bool PredictTheWinner(vector<int>& nums){int dp[25][25];memset(dp,0,sizeof(dp));for(int tail=1; tail<nums.size(); tail++)for(int head=tail; head>=0; head--){int get_head = nums[head] - dp[head+1][tail];int get_tail = nums[tail] - dp[head][tail-1];dp[head][tail] = max(get_head, get_tail);}return dp[0][nums.size()-1]>=0;}
};

转载于:https://www.cnblogs.com/jasonlixuetao/p/10545715.html

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