基本题目

509. 斐波那契数

f(n) = f(n-1) + f(n-2)，滚动数组优化

class Solution {
public:
    int fib(int n) {
        if (n == 0) return 0;
        int a = 0, b = 1, tmp;
        for (int i = 1; i < n; i++) {
            tmp = a + b;
            a = b;
            b = tmp;
        }
        return b;
    }
};

还可以使用通项公式、矩阵快速幂计算：

70. 爬楼梯

与斐波那契完全等价

class Solution {
public:
    int climbStairs(int n) {
        int a = 0, b = 1, tmp;
        for (int i = 0; i < n; i++) {
            tmp = a + b;
            a = b;
            b = tmp;
        }
        return b;
    }
};

746. 使用最小花费爬楼梯

f(n) = min(f(n-1)+c[n-1], f(n-2)+c[n-2])，滚动数组优化

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        int a = 0, b = 0, tmp;
        for (int i = 1; i < n; i++) {
            tmp = min(a + cost[i - 1], b + cost[i]);
            a = b;
            b = tmp;
        }
        return b;
    }
};

62. 不同路径

f(i,j) = f(i-1,j) + f(i,j-1)，滚动行优化

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<int> row(n, 1);
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                row[j] += row[j - 1];
            }
        }
        return row.back();
    }
};

63. 不同路径 II

f(i,j) = f(i-1,j) + f(i,j-1) or 0，滚动行优化

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<int> row(n);
        row[0] = obstacleGrid[0][0] ? 0 : 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (obstacleGrid[i][j]) row[j] = 0;
                else if (j > 0) row[j] += row[j - 1];
            }
        }
        return row.back();
    }
};

343. 整数拆分

将每个整数所有可能的拆分情况遍历，取最大的构造 dp

class Solution {
public:
    int integerBreak(int n) {
        vector<int> maxProduct(n);
        for (int i = 1; i <= n; i++) {
            int maxProd = 1;
            for (int j = 1; j <= i / 2; j++) {
                maxProd = max(maxProd, maxProduct[j] * maxProduct[i - j]);
            }
            if (i == n) return maxProd;
            maxProduct[i] = max(maxProd, i);
        }
        return 0;
    }
};

注意到，对于任意大于等于 4 的数 x，将其拆分为 2 与 x - 2 后得到的乘积一定大于 x。因此在最终的拆分结果中，一定不包含大于等于 4 的数。此外，拆分结果中也一定不包含 1。

class Solution {
public:
    int integerBreak(int n) {
        if (n <= 3) return n - 1;

        int m3 = 1, m2 = 2, m1 = 3;
        for (int i = 4; i <= n; i++) {
            int maxProd = max(2 * m2, 3 * m3);
            if (i == n) return maxProd;
            m3 = m2;
            m2 = m1;
            m1 = max(maxProd, i);
        }
        return 0;
    }
};

96. 不同的二叉搜索树

搜索树的形状数与二叉树的总形状数一样，都等于卡特兰数

class Solution {
public:
    int numTrees(int n) {
        vector<int> dp(n + 1, 1);
        for (int i = 2; i <= n; i++) {
            int sum = 0;
            for (int j = 0; j < i; j++) {
                sum += dp[j] * dp[i - j - 1];
            }
            dp[i] = sum;
        }
        return dp[n];
    }
};