代码随想录算法训练营第34天 | 第九章 动态规划07

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今日记录

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198.打家劫舍

leetcode链接

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() + 1, 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 - 2] + nums[i], dp[i - 1]);
        }
        return dp[nums.size() - 1];
    }
};

213.打家劫舍Ⅱ

Leetcode链接

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 0)
            return 0;
        if (nums.size() == 1)
            return nums[0];
        int result1 = robRange(nums, 0, nums.size() - 2); // 包含首元素
        int result2 = robRange(nums, 1, nums.size() - 1); // 包含尾元素
        return max(result1, result2);
    }

    int robRange(vector<int>& nums, int start, int end) {
        if (end == start)
            return nums[start];
        vector<int> dp(nums.size());
        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 - 2] + nums[i], dp[i - 1]);
        }
        return dp[end];
    }
};

337.打家劫舍Ⅲ

Leetcod链接

class Solution {
public:
    int rob(TreeNode* root) {
        vector<int> result = robTree(root);
        return max(result[0], result[1]);
    }
    vector<int> robTree(TreeNode* cur) {
        //后序遍历：左右中
        if (cur == NULL) return vector<int>{0, 0};//初始化为0：实际0不偷，1偷
        vector<int> left = robTree(cur->left);
        vector<int> right = robTree(cur->right);
        // 偷cur不能偷左右节点。
        //中
        int val1 = cur->val + left[0] + right[0];
        // 不偷cur取偷or不偷左右节点较大的情况
        int val2 = max(left[0], left[1]) + max(right[0], right[1]);
        return {val2, val1};
    }
};

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总结