Leetcode:375. Guess Number Higher or Lower II
2018-01-09 14:15
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Description
We are playing the Guess Game. The game is as follows:I pick a number from 1 to n. You have to guess which number I picked.
Every time you guess wrong, I’ll tell you whether the number I picked is higher or lower.
However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.
Example:
n = 10, I pick 8.
First round: You guess 5, I tell you that it’s higher. You pay ¥5.
Second round: You guess 7, I tell you that it’s higher. You pay ¥7.
Third round: You guess 9, I tell you that it’s lower. You pay¥9.
Game over. 8 is the number I picked.
You end up paying ¥5 + ¥7 + ¥9 = ¥21.
Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.
解题思路
本题大意就是让你猜一个数,猜错要罚钱,同时会有提示你猜大了还是小了,极小化极大算法解决。算法分析如下:
dp[i][j]表示从数字i到j之间猜中任意一个数字最少需要花费的钱数
1、遍历每一段区间[j, i],维护一个全局最小值global_min变量,
2、遍历该区间中的每一个数字,计算局部最大值local_max = k + max(dp[j][k - 1], dp[k + 1][i]),将该区间在每一个位置都分为两段,然后取当前位置的花费加上左右两段中较大的花费之和为局部最大值(取最坏的情况)
3、然后更新全局最小值,最后在更新dp[j][i]的时候看j和i是否是相邻的,相邻的话赋为i,否则赋为global_min
具体分析如下:
1、只有一个数字,cost为0。
2、有两个数字,比如1和2,猜1,即使不对,cost也比猜2要低。
3、有三个数字1,2,3,那么先猜2,根据对方的反馈,就可以确定正确的数字,所以cost最低为2。
4、有四个数字1,2,3,4,那么用k来遍历所有的数字,然后再根据k分成的左右两个区间,取其中的较大cost加上k。
当k为1时,左区间为空,所以cost为0,而右区间2,3,4,根据之前的分析应该取3,所以整个cost就是1+3=4。
当k为2时,左区间为1,cost为0,右区间为3,4,cost为3,整个cost就是2+3=5。
当k为3时,左区间为1,2,cost为1,右区间为4,cost为0,整个cost就是3+1=4。
当k为4时,左区间1,2,3,cost为2,右区间为空,cost为0,整个cost就是4+2=6。
综上k的所有情况,此时应该取整体cost最小的,即4,为最后的答案,这就是极小化极大算法
class Solution { public: int getMoneyAmount(int n) { vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));//注意一定要先赋值为0 for (int i = 2; i <= n; ++i) { for (int j = i - 1; j > 0; --j) { int global_min = INT_MAX; for (int k = j + 1; k < i; ++k) { int local_max = k + max(dp[j][k - 1], dp[k + 1][i]); global_min = min(global_min, local_max); } dp[j][i] = j + 1 == i ? j : global_min; } } return dp[1] ; } };
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