zoukankan      html  css  js  c++  java
  • LC 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. 1 <= length of the array <= 20.
    2. Any scores in the given array are non-negative integers and will not exceed 10,000,000.
    3. If the scores of both players are equal, then player 1 is still the winner.
     
     
    Runtime: 8 ms, faster than 19.93% of C++ online submissions for Predict the Winner.
    Memory Usage: 5.1 MB, less than 0.75% of C++ online submissions for Predict the Winner.
     
     
    class Solution {
    public:
      bool PredictTheWinner(vector<int>& nums) {
        int N = nums.size();
    
        vector<vector<int>> mtx(N+1, vector<int>(N+1, 0));
    
        for(int gap = 0; gap < nums.size(); gap++) {
          for(int i=1; i<nums.size()+1; i++) {
            for(int j=i; j <= i+gap && j <= N; j++) {
              if(j == i) mtx[i][j] = nums[i-1];
              else if(j == i+1) {
                mtx[i][j] = max(nums[i-1], nums[j-1]);
                mtx[j][i] = min(nums[i-1], nums[j-1]);
              }
              else {
                int takeleft = mtx[j][i+1] + nums[i-1];
                int takeright = mtx[j-1][i] + nums[j-1];
                mtx[i][j] = max(takeleft, takeright);
                if(takeleft > takeright) mtx[j][i] = mtx[i+1][j];
                else if(takeleft < takeright) mtx[j][i] = mtx[i][j-1];
                else mtx[j][i] = min(mtx[i+1][j], mtx[i][j-1]);
              }
            }
          }
        }
        bool canwin = mtx[1][N] >= mtx[N][1];
        return canwin;
      }
    };
  • 相关阅读:
    Linux启动报错missing operating system
    Linux tmp目录自动清理总结
    ORACLE口令管理
    oom_kill_process造成数据库挂起并出现found dead shared server
    Oracle shutdown immediate遭遇ORA-24324 ORA-24323 ORA-01089
    从flink-example分析flink组件(1)WordCount batch实战及源码分析
    从mysql高可用架构看高可用架构设计
    由mysql分区想到的分表分库的方案
    六边形效果
    迷你MVVM框架 avalonjs1.5.2 发布
  • 原文地址:https://www.cnblogs.com/ethanhong/p/10354419.html
Copyright © 2011-2022 走看看