In the "100 game," two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins. What if we change the game so that players cannot re-use integers? For examp…

动态规划是一种top-down求解模式,关键在于分解和求解子问题,然后根据子问题的解不断向上递推,得出最终解 因此dp涉及到保存每个计算过的子问题的解,这样当遇到同样的子问题时就不用继续向下求解而直接可以得到结果.状态压缩就是用来保存子问题的解的,主要思想是把所有可能的状态(子问题)用一个数据结构(通常是整数)统一表示,再用map把每个状态和对应结果关联起来,这样每次求解子问题时先find一下,如果map里面已经有该状态的解就不用再求了:同样每次求解完一个状态的解后也要将其放入map中保存 状态…

In the "100 game," two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins. What if we change the game so that players cannot re-use integers? For examp…

详见:https://leetcode.com/problems/can-i-win/description/ C++: class Solution { public: bool canIWin(int maxChoosableInteger, int desiredTotal) { if (maxChoosableInteger >= desiredTotal) { return true; } if (maxChoosableInteger * (maxChoosableInteger +…

原题链接 两个人依次从1~maxNum中选取数字(不可重复选取同一个),累和.当一方选取数字累和后结果大于等于给定的目标数字,则此人胜利. 题目给一个maxNum和targetNum,要求判断先手能否胜利. 思路: 首先判断两种特殊条件: 可选最大值大于等于目标值,直接返回true. 其中一个人可选的最大值小于目标值,直接返回false. 具体再看题目,题目给出maxNum不大于20,则可状态压缩为一个int(32位)来保存每个数字是否被使用过(题解中利用1-32位进行存储). 利用一个map存…

In the "100 game," two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins. What if we change the game so that players cannot re-use integers? For examp…

我能赢吗 在 "100 game" 这个游戏中,两名玩家轮流选择从 1 到 10 的任意整数,累计整数和,先使得累计整数和达到 100 的玩家,即为胜者. 如果我们将游戏规则改为 "玩家不能重复使用整数" 呢? 例如,两个玩家可以轮流从公共整数池中抽取从 1 到 15 的整数(不放回),直到累计整数和 >= 100. 给定一个整数 maxChoosableInteger (整数池中可选择的最大数)和另一个整数 desiredTotal(累计和),判断先出手的玩…

https://leetcode.com/problems/can-i-win/description/ In the "100 game," two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins. What if we change the g…

源代码地址:https://github.com/hopebo/hopelee 语言:C++ 301. Remove Invalid Parentheses Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results. Note: The input string may contain letters other tha…

Contest 91 (2018年10月24日,周三) 链接:https://leetcode.com/contest/weekly-contest-91/ 模拟比赛情况记录:第一题柠檬摊的那题6分钟AC,然后是第二题树的距离K的结点那题比较久,大概写了30分钟,第三题翻转矩阵那题第一次提交错误了,列的优化方法思路错了,WA.后来比赛时间到了才改过来.最后一题最短子数组的和比K大的这题不会做. [860]Lemonade Change   (第一题) 一个柠檬摊,有一队人排队要买柠檬,一个5刀,…