198. 213. 337. House Robber -- 不取相邻值的最大值

198. House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size(), i;
        if( == n)
            return ;
        if( == n)
            return nums[];
        vector<int> dp(n);
        dp[] = nums[];
        dp[] = max(nums[], nums[]);
        for(i = ; i < n; i++)
        {
            dp[i] = max(dp[i-], dp[i-] + nums[i]);
        }
        return dp[n-];
    }
};

213. House Robber II

After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

class Solution {

public:

    int orginal_rob(vector<int> &money, int start, int end) {

        int n2=; 

        int n1=; 

        for (int i=start; i<end; i++){

            int current = max(n1, n2 + money[i]);

            n2 = n1;

            n1 = current;

        }

        return n1;

    }

    int rob(vector<int>& nums) {

        int n = nums.size();

        switch (n) {

            case :

                return ;

            case :

                return nums[];

            case :

                return max(nums[], nums[]);

            default:

                /*

                 * the idea is we cannot rob[0] and rob[n-1] at same time

                 * so, we rob [0 .. n-2] or [1 .. n-1], can return the maxinum one.

                 */

                int m1 = orginal_rob(nums, , n-);

                int m2 = orginal_rob(nums, , n);

                return max(m1, m2);

        }

    }

};

337. House Robber III

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

Example 1:

     3
    / \
   2   3
    \   \
     3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

     3
    / \
   4   5
  / \   \
 1   3   1

Maximum amount of money the thief can rob = 4 + 5 = 9.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    map<TreeNode*, int> m;
public:
    int rob(TreeNode* root) {
        if(root == NULL)
            return ;
        if(m.find(root) != m.end())
            return m[root];
        int left = rob(root->left);
        int right = rob(root->right);
        int child = left + right;
        int ans = root->val;
        if(root->left)
        {
            ans += rob(root->left->left) + rob(root->left->right);
        }
        if(root->right)
        {
            ans += rob(root->right->left) + rob(root->right->right);
        }
        m[root] = max(ans, child);
        return m[root];
    }
};