lintcode Guess Number Game II

动态规划解析:

解法一: 记忆化搜索

1. state: dp[i][j] 为 从i 到 j 最少消耗的钱2. function: dp[i][j] = min(value[k] + max(dp[i][k],dp[k+1][j]))3. initialize: dp[0..n][0..n] = -14. answer: dp[0][n-1]详细代码: c++

    int** dp;     int getMoneyAmountHelper (int low, int high) {        if (low >= high) {            return 0;        }        if (dp[low][high] != -1) {            return dp[low][high];        }        int maxCost = pow(2,31) - 1;        for (int i = low; i <= high; i++){            int left = getMoneyAmountHelper(low,i-1);            int right = getMoneyAmountHelper(i+1, high);            maxCost = min(maxCost, max(left, right));        }        return dp[low][high] = max1;    }     int getMoneyAmount(int n) {       dp = new int*[n];       for(int i = 0; i < n; i++) {           dp[i] = new int[n];           for (int j = 0; j < n; j++) {               dp[i][j] = -1;           }       }       return getMoneyAmountHelper(0,n-1);    }

解法二: 循环

和记忆化搜索的差别在与 要注意初始化顺序

int getMoneyAmount(int n) {    int dp[n][n];    const int INTMAX = pow(2, 31) - 1;    for (int i = 0; i < n; i++) {        for (int j = 0; j < n; j++) {            dp[i][j] = INTMAX;        }    }    for (int i = 0; i < n; i++) {        dp[i][i] = 0;    }    for (int width = 1; width < n; width++) {        for (int i = 0, j = i + width; j < n; i++, j++) {            for (int k = i; k <= j; ++k) {                if (k > i && k < j) {                    dp[i][j] = min(dp[i][j], k + 1 +                     max(dp[i][k - 1], dp[k + 1][j]));                } else if (k == i) {                    dp[i][j] = min(dp[i][j], k + 1 + dp[i + 1][j]);                } else {                    dp[i][j] =min(dp[i][j], k + 1 + dp[i][j-1]);                }            }        }    }    return dp[0][n - 1];}