8. leetcode(动态规划-打家劫舍+股票问题)

打家劫舍

1. leetcode 198 打家劫舍

//1. leetcode 198 打家劫舍

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        }
        if (nums.size() == 1) {
            return nums[0];
        }

        vector<int> dp(nums.size(), 0);

        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);

        for (int i = 2; i < nums.size(); i++) {
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        return dp[nums.size() - 1]; 
    }
};

2. leetcode 213 打家劫舍 II

//2. leetcode 213 打家劫舍 II

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        }
        if (nums.size() == 1) {
            return nums[0];
        }
        //包含首 舍弃尾
        int res1 = robRange(nums, 0, nums.size() - 2);
        //包含尾 舍弃首
        int res2 = robRange(nums, 1, nums.size() - 1);
        return max(res1, res2);
    }

    int robRange(vector<int>& nums, int start, int end) {
        if (start == end) {
            return nums[start];
        }

        vector<int> dp(nums.size(), 0);
        dp[start] = nums[start];
        dp[start + 1] = max(nums[start], nums[start + 1]);

        for (int i = start + 2; i <= end; i++) {
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);
        }

        return dp[end];
    }
};

3. leetcode 337 打家劫舍 III

//3. leetcode 337 打家劫舍 III


/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int rob(TreeNode* root) {
        vector<int> res = robTree(root);
        return max(res[0], res[1]);
    }
    //长度为2 的数组， 0不偷   1偷
    vector<int> robTree(TreeNode* node) {
        if (node == nullptr) {
            return vector<int>{0, 0};
        }

        vector<int> left = robTree(node->left);
        vector<int> right = robTree(node->right);
        //偷node
        int val1 = node->val + left[0] +right[0];
        //不偷node
        int val2 = max(left[0], left[1]) + max(right[0], right[1]);

        return {val2, val1};
    }
};

股票系列

1. leetcode 121 买卖股票的最佳时机

//1. leetcode 121 买卖股票的最佳时机

//暴力解法

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0;
        for (int i = 0; i < prices.size(); i++) {
            for (int j = i + 1; j < prices.size(); j++) {
                res = max(res, prices[j] - prices[i]);
            }
        }
        return res;
    }
};

//单调栈
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0;
        vector<int> st;
        prices.emplace_back(-1);  //哨兵
        for (int i = 0; i < prices.size(); i++) {
            while (!st.empty() && st.back() > prices[i]) {  //维护单调栈
                res = max(res, st.back() - st.front());  //维护最大值
                st.pop_back();
            }
            st.emplace_back(prices[i]);
        }
        return res;
    }
};


//动态规划
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        if (n == 0) return 0;
        int minprice = prices[0];
        vector<int> dp(n, 0);

        for (int i = 1; i < n; i++) {
            minprice = min(minprice, prices[i]);
            dp[i] = max(dp[i - 1], prices[i] - minprice);
        }
        return dp[n - 1];
    }
};

//一次遍历  贪心

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int minprice = INT_MAX;
        int maxProfit = 0;
        for (auto price : prices) {
            maxProfit = max(maxProfit, price - minprice);
            minprice = min(minprice, price);
        }
        return maxProfit;
    }
};

2. leetcode 122 买卖股票的最佳时机 II

//2.leetcode 122 买卖股票的最佳时机 II

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int len  = prices.size();
        vector<vector<int>> dp(len, vector<int>(2, 0));
        dp[0][0] -= prices[0];
        dp[0][1] = 0;

        for (int i = 1; i < len; i++) {
            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i]);
        }
        return dp[len - 1][1];
    }
};


class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int len = prices.size();
        //开辟一个2*2的数组
        vector<vector<int>> dp(2, vector<int>(2));
        dp[0][0] -= prices[0];
        dp[0][1] = 0;
        for (int i = 1; i < len; i++) {
            dp[i % 2][0] = max(dp[(i - 1) % 2][0], dp[(i - 1) % 2][1] - prices[i]);
            dp[i % 2][1] = max(dp[(i - 1) % 2][1], prices[i] + dp[(i - 1) % 2][0]);
        }
        return dp[(len - 1) % 2][1];
    }
};

3. leetcode 123 买卖股票的最佳时机 III

//3. leetcode 123 买卖股票的最佳时机 III

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0) return 0;
        vector<vector<int>> dp(prices.size(), vector<int>(5, 0));
        dp[0][1] = -prices[0];
        dp[0][3] = -prices[0];

        for (int i = 1; i < prices.size(); i++) {
            dp[i][0] = dp[i - 1][0];
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
            dp[i][2] = max(dp[i - 1][2], dp[i - 1][1] + prices[i]);
            dp[i][3] = max(dp[i - 1][3], dp[i - 1][2] - prices[i]);
            dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i]);

        }
        return dp[prices.size() - 1][4];
    }
};



class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() == 0) return 0;
        vector<int> dp(5, 0);
        dp[1] = -prices[0];
        dp[3] = -prices[0];

        for (int i = 1; i < prices.size(); i++) {
            dp[1] = max(dp[1], dp[0] - prices[i]);
            dp[2] = max(dp[2], dp[1] + prices[i]);
            dp[3] = max(dp[3], dp[2] - prices[i]);
            dp[4] = max(dp[4], dp[3] + prices[i]);
        }
        return dp[4];
    }
};

4. leetcode 188 买卖股票的最佳时机 IV

//4. leetcode 188 买卖股票的最佳时机 IV
class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        if (prices.size() == 0) return 0;
        vector<vector<int>> dp(prices.size(), vector<int>(2 * k + 1, 0));
        for (int j = 1; j < 2 * k; j += 2) {
            dp[0][j] = -prices[0];
        }

        for (int i = 1; i < prices.size(); i++) {
            for (int j = 0; j < 2 * k - 1; j += 2) {
                dp[i][j + 1] = max(dp[i - 1][j + 1], dp[i - 1][j] - prices[i]);
                dp[i][j + 2] = max(dp[i - 1][j + 2], dp[i - 1][j + 1] + prices[i]);
            }
        }
        return dp[prices.size() - 1][2 * k];
    }
};

5. leetcode 309 最佳买卖股票时机含冷冻期

//5. leetcode 309 最佳买卖股票时机含冷冻期

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        if (n == 0) return 0;
        vector<vector<int>> dp(n, vector<int>(4, 0));
        dp[0][0] -= prices[0];
        for (int i = 1; i < n; i++) {
            dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3], dp[i - 1][1]) - prices[i]);
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][3]);
            dp[i][2] = dp[i - 1][0] + prices[i];
            dp[i][3] = dp[i - 1][2];
        }
        return max(dp[n - 1][3], max(dp[n - 1][1], dp[n - 1][2]));
    }
};

6. leetcode 714 买卖股票的最佳时机含手续费

//6. leetcode 714 买卖股票的最佳时机含手续费

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2, 0));
        dp[0][0] -= prices[0];
        for (int i = 1; i < n; i++) {
            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee);
        }
        return max(dp[n - 1][0], dp[n - 1][1]);
    }
};