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> f(nums.size()+10,0);f[1]=nums[0];f[2]=max(nums[0],nums[1]);for(int i=3;i<=nums.size();i++){f[i]=max(f[i-2]+nums[i-1],f[i-1]);}return f[nums.size()];}
};

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];int res1 = robRange(nums,1,nums.size()-1);int res2 = robRange(nums,0,nums.size()-2);return max(res1,res2);}int robRange(vector<int>& nums,int st,int ed) {if (ed == st) return nums[st];vector<int> f(nums.size()+1,0);f[st]=nums[st];f[st+1]=max(nums[st],nums[st+1]);for(int i=st+2;i<=ed;i++){f[i]=max(f[i-2]+nums[i],f[i-1]);}return f[ed];}
};

337. 打家劫舍 III

class Solution {
public:int rob(TreeNode* root) {vector<int> result = robTree(root);return max(result[0], result[1]);}// 长度为2的数组，0：不偷，1：偷vector<int> robTree(TreeNode* cur) {if (cur == NULL) return vector<int>{0, 0};vector<int> left = robTree(cur->left);vector<int> right = robTree(cur->right);// 偷cur，那么就不能偷左右节点。int val1 = cur->val + left[0] + right[0];// 不偷cur，那么可以偷也可以不偷左右节点，则取较大的情况int val2 = max(left[0], left[1]) + max(right[0], right[1]);return {val2, val1};}
};