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.
猜数字,猜错了就得付与所猜的数目一致的钱,最后求出保证你能赢的钱数(即求出保证你获胜所花费最少的钱数)
动态规划,嗯,挺难的我觉得。。。
在1-n个数里面,我们任意猜一个数(设为i),保证获胜所花的钱应该为 i + max(w(1 ,i-1), w(i+1 ,n)),这里w(x,y)表示猜范围在(x,y)的数保证能赢应花的钱,则我们依次遍历 1-n作为猜的数,求出其中的最小值即为答案。
时间复杂度O(n^2),空间复杂度O(n^2)
class Solution {
public:
int getMoneyAmount(int n) {
vector > dp(n+1, vector(n+1, 0));
if (n <= 1) return 0;
return cost(dp, 1, n);
}
int cost(vector >& dp, int st, int ed) {
if (st >= ed) return 0;
if (dp[st][ed]) return dp[st][ed];
int res = INT_MAX;
for (int i = st; i <= ed; ++i) {
res = min(res, i + max(cost(dp, st, i-1), cost(dp, i+1, ed)));
}
dp[st][ed] = res;
return res;
}
};
碎碎念的分析:(懂了的不用看)
设w为所需要付的钱,当n=1时,只有一个数肯定能赢,w=0;当n=2时,我们可能猜的数为1或2,选择猜1,这时正确答案是1时不用花钱,是2时只需要花1,猜2时同理,比较一下这两种情况,我们选择猜1就能保证获胜且花费最小,w=1; 当n=3时,我们选猜2,最多只需花2就能保证获胜,w=2,这几种是帮助理解的初始状态。
当n很大时,我们随便猜一个数m(m