day33

打家劫舍
class Solution {
public:
int rob(vector& nums) {
if (nums.size() == 0) return 0;
if (nums.size() == 1) return nums[0];
vector 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-2] + nums[i], dp[i-1]);
}
return dp[nums.size() - 1];
}
};

打家劫舍II

class Solution {
public:
int rob1(vector& nums) {
if (nums.size() == 0) return 0;
if (nums.size() == 1) return nums[0];
vector 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-2] + nums[i], dp[i-1]);
}
return dp[nums.size() - 1];
}
int rob(vector& nums) {
if (nums.size() == 0) return 0;
if (nums.size() == 1) return nums[0];
vector vec1;
for(int i = 0; i < nums.size() - 1; ++i)
{
vec1.push_back(nums[i]);
}
vector vec2;
for(int i = 1; i < nums.size(); ++i)
{
vec2.push_back(nums[i]);
}
return max(rob1(vec1), rob1(vec2));
}
};

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:
  vector traversal(TreeNode root)
  {
  if(root == nullptr)
  {
  return {0, 0};
  }
  vector left = traversal(root->left);
  vector right = traversal(root->right);
  int val1 = root->val + left[0] + right[0];
  int val2 = max(left[0], left[1]) + max(right[0], right[1]);
  return {val2, val1};
  }
  int rob(TreeNode root) {
  vector ret = traversal(root);
  return max(ret[0], ret[1]);

  }
  };