374. Guess Number Higher or Lower, 375. Guess Number Higher or Lower II

374. Guess Number Higher or Lower

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.

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/* The guess API is defined in the parent class GuessGame.
   @param num, your guess
   @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
      int guess(int num); */

public class Solution extends GuessGame {
    public int guessNumber(int n) {
        int left = 1, right = n;
        
        while (left <= right) {
            int mid = (right - left) / 2 + left;
            int rt = guess(mid);
            if (rt == 0) return mid;
            if (rt > 0) {
                left = mid + 1;
            }else {
                right = mid - 1;
            }
        }
        
        return left;
    }
}

375. Guess Number Higher or Lower II

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.

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题意有点绕：找一个花费最小的通解
brute force的算法是枚举所有的可能性，找出最小（最优）的解。
给定[start, end], 猜i点之后的cost是cost[i] = i + Math.max(cost[start, i - end], cost[i + 1, end]), 取2者中间最大是因为不能确定最终结果在左还是在右
ref: https://leetcode.com/problems/guess-number-higher-or-lower-ii/solution/

Solution #1 DP 是对brute force的优化
重点：这题的2维DP是按对角线来填充，因为[len]是基于[len - i]...[len - 1］
以后写dp要理解其本身的通项公式
O(n^3)

class Solution {
    public int getMoneyAmount(int n) {
        int[][] dp = new int[n + 1][n + 1];
        
        for (int len = 2; len <= n; len++) {
            for (int start = 1; start < n - len + 2; start++) {
                int min = Integer.MAX_VALUE;
                for (int i = start; i <= start + len - 1; i++) {
                    
                    int v = 0;
                    if (i == start + len - 1) {
                        v = i + dp[start][i - 1];
                    }else 
                        v = i + Math.max(dp[start][i - 1], dp[i + 1][start + len - 1]);
                    
                    min = Math.min(min, v);
                }
                dp[start][start + len - 1] = min;
            }
        }
        
        return dp[1][n];
    }
}

进一步优化: [start, (start + end) / 2] 永远比 [(start + end) / 2, end]小，所以只需计算后半部分。BigO不变