Predict the Winner

Given an array of scores that are non-negative integers. Player 1 picks one of the numbers from either end of the array followed by the player 2 and then player 1 and so on. Each time a player picks a number, that number will not be available for the next player. This continues until all the scores have been chosen. The player with the maximum score wins.

Given an array of scores, predict whether player 1 is the winner. You can assume each player plays to maximize his score.

Solution 1

1st version 

丑陋

class Solution {    int a1,a2;public:    bool PredictTheWinner(vector<int>& nums) {        return PredictTheWinnerScoreFirst(nums,0,nums.size()-1,0,0);    }        bool PredictTheWinnerScoreFirst(vector<int>& nums,int l, int h, int a1, int a2) {        int s = h-l+1;        int t;        bool r;        if(1==s){            return a1+nums[l]>=a2;        }                if(2==s){            t = abs(nums[l] - nums[h]);            return a1+t>=a2;        }        //1. choose low        r = PredictTheWinnerScoreSecond(nums, l+1, h, a1+nums[l], a2);        if(r == false) return true;                //2. choose high        r = PredictTheWinnerScoreSecond(nums, l, h-1, a1+nums[h], a2);         if(r == false) return true;                return false;    }        bool PredictTheWinnerScoreSecond(vector<int>& nums,int l, int h, int a1, int a2) {        int s = h-l+1;        int t;        bool r;        if(1==s){            return a2+nums[l]>a1;        }                if(2==s){            t = abs(nums[l] - nums[h]);            return a2+t>a1;        }                //1. choose low        r = PredictTheWinnerScoreFirst(nums, l+1, h, a1, a2+nums[l]);        if(r == false) return true;                //2. choose high        r = PredictTheWinnerScoreFirst(nums, l, h-1, a1, a2+nums[h]);         if(r == false) return true;                return false;            }};

优化：2nd version  

进一步抽象，删除if(2--s)

class Solution {    int a1,a2;public:    bool PredictTheWinner(vector<int>& nums) {        return PredictTheWinnerScoreFirst(nums,0,nums.size()-1,0,0);    }        bool PredictTheWinnerScoreFirst(vector<int>& nums,int l, int h, int a1, int a2) {        int s = h-l+1;        int t;        bool r;        if(1==s){            return a1+nums[l]>=a2;        }                //1. choose low        r = PredictTheWinnerScoreSecond(nums, l+1, h, a1+nums[l], a2);        if(r == false) return true;                //2. choose high        r = PredictTheWinnerScoreSecond(nums, l, h-1, a1+nums[h], a2);         if(r == false) return true;                return false;    }        bool PredictTheWinnerScoreSecond(vector<int>& nums,int l, int h, int a1, int a2) {        int s = h-l+1;        int t;        bool r;        if(1==s){            return a2+nums[l]>a1;        }                //1. choose low        r = PredictTheWinnerScoreFirst(nums, l+1, h, a1, a2+nums[l]);        if(r == false) return true;                //2. choose high        r = PredictTheWinnerScoreFirst(nums, l, h-1, a1, a2+nums[h]);         if(r == false) return true;                return false;            }};

优化：3rd version 

PredictTheWinnerScoreFirst 和 PredictTheWinnerScoreSecond 合并

class Solution {public:    bool PredictTheWinner(vector<int>& nums) {        return helper(nums,0,nums.size()-1,0,true);    }        bool helper(vector<int>& nums,int l, int h, int t, bool isFirst) {        bool r;        if(l==h){            return isFirst? t+nums[l]>=0 : nums[l]>t;        }                //1. choose low        isFirst? t+=nums[l] : t-=nums[l];        r = helper(nums, l+1, h, t,!isFirst);        if(r == false) return true;                //2. choose high        isFirst? t+=nums[h] : t-=nums[h];        r = helper(nums, l, h-1, t,!isFirst);         if(r == false) return true;                return false;    }}；

Solution 2

1st version 

class Solution {    int a1,a2;public:    bool PredictTheWinner(vector<int>& nums) {        return helper(nums,0,nums.size()-1)>=0;    }        int helper(vector<int>& nums,int l, int h) {        int r1;        int r2;        if(l==h){            return nums[l];        }                //1. choose low        r1 = nums[l]-helper(nums,l+1,h);        //2. choose high        r2 = nums[h] - helper(nums, l, h-1);                         return max(r1,r2);    }};

2nd version 增加记忆

class Solution {    int t[20][20];    bool b[20][20];public:    bool PredictTheWinner(vector<int>& nums) {        for(int i=0; i<20; i++){            for(int j=0; j<20; j++){                b[i][j] = false;            }        }        return helper(nums,0,nums.size()-1)>=0;    }        int helper(vector<int>& nums,int l, int h) {        if(b[l][h]) return t[l][h];                int r1;        int r2;        if(l==h){            return nums[l];        }                //1. chose low        r1 = nums[l]-helper(nums,l+1,h);        //2. chose high        r2 = nums[h] - helper(nums, l, h-1);                 b[l][h] = true;         t[l][h] =  max(r1,r2);        return  t[l][h];    }};

最坏复杂度

W(n) = 2W(n-1)

W(n-1) = 2^2W(n-2)

W(2) = W(n-(n-2)) = 2^（n-1）W（1） = 2^(n-1)