算法分析與設計課程作業第十八週#1

算法分析與設計課程作業第十八週#1

選了道題做來保持下感覺，複習周還真是忙呀。

486. 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.
Example 1:
Input: [1, 5, 2]
Output: False
Explanation: Initially, player 1 can choose between 1 and 2.
If he chooses 2 (or 1), then player 2 can choose from 1 (or 2) and 5. If player 2 chooses 5, then player 1 will be left with 1 (or 2).
So, final score of player 1 is 1 + 2 = 3, and player 2 is 5.
Hence, player 1 will never be the winner and you need to return False.

Example 2:
Input: [1, 5, 233, 7]
Output: True
Explanation: Player 1 first chooses 1. Then player 2 have to choose between 5 and 7. No matter which number player 2 choose, player 1 can choose 233.
Finally, player 1 has more score (234) than player 2 (12), so you need to return True representing player1 can win.

Note:
1 <= length of the array <= 20.
Any scores in the given array are non-negative integers and will not exceed 10,000,000.
If the scores of both players are equal, then player 1 is still the winner.

思路

就是無論對手還是自己都會盡力想讓自己得到最多的分（同時，讓對手得到最少的分，兩個是一個意思）。
可以用dp[i][j]表示能從i到j的數組元素中拿最多能拿多少分，具體狀態轉移式可見代碼。

代碼

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        vector<vector<int>> dp(nums.size(), vector<int>(nums.size(), 0));
        for(int l = 1; l <= nums.size(); l++){
            for(int i = 0; i <= nums.size() - l; i++){
                if(l == 1){
                    dp[i][i+l-1] = nums[i];
                }
                else if(l == 2){
                    dp[i][i+l-1] = max(nums[i], nums[i+l-1]);
                }
                else{
                    int chooseleft = nums[i] + min(dp[i+1][i+l-2],dp[i+2][i+l-1]);
                    int chooseright = nums[i+l-1] + min(dp[i][i+l-3],dp[i+1][i+l-2]);
                    dp[i][i+l-1] = max(chooseleft, chooseright);
                }

            }
        }
        int sum = 0;
        for(int i = 0; i < nums.size(); i++){
            sum += nums[i];
        }
        return dp[0][nums.size()-1] >= sum/2 + sum%2;
    }
};