leetcode numberguess size 1/2 (two points 374, 375, dynamic programming interval)

374 Guess Size

Topic Description: We're playing a guessing game. Rules of the game are as follows:
I n select a number from 1 to. I guess you need to choose which number. Every time you guess wrong, I'll tell you this figure is big or small. You call a pre-defined interfaces guess (int num), it will return three possible outcomes (-1, 1 or 0): -1: I figure is relatively small; 1: I figure larger; 0: Congratulations! you guessed right!
Ideas: half guessing.

// Forward declaration of guess API.
// @param num, your guess
// @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
int guess(int num);

int ABC(int n);

class Solution {
public:
    int guessNumber(int n) {
        long long low = 1, high = n, mid = 0;
        int ans = 0;
        while(low <= high) {
            mid = (low + high) / 2;
            if(guess(mid) <= 0) {
                ans = mid;
                high = mid - 1;
            }
            else
                low = mid + 1;
        }
        return ans;
    }
};

375 Guess Size 2

Topic Description: We're playing a guessing game, rules of the game are as follows:
I select a number from 1 to n, you guess what number I chose. Every time you guess wrong, I will tell you, I chose numbers than your big or small. However, when you guess wrong as digital x and when you need to pay x amount of cash. I guess until you choose the numbers, you win the game be considered.
Given n ≥ 1, at least you need to calculate how much cash to make sure you can win this game.
Ideas: dynamic programming range.
Topic understand: give you a n, he may have the actual numbers (ie I choose the number) is any 1-n one (denoted by i), for each i, there are many guess law, there is an answer spending is the smallest. We have a minimum cost for each i, then the answer is the maximum of all i corresponding minimum cost.

class Solution {
public:
    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n+2, vector<int>(n+2, 100000000));
        for(int i = 1; i <= n; i++) {
            for(int j = 0; j <= i; j++){
                dp[i][j] = 0;
            }
        }
        for(int len = 2; len <= n; len++) {
            for(int L = 1; L + len - 1 <= n; L++) {
                int R = L + len - 1;
                for(int mid = L; mid <= R; mid++) 
                    dp[L][R] = min(dp[L][R], max(dp[L][mid - 1], dp[mid + 1][R]) + mid); 
            }
        }
        return dp[1][n];
    }
};