class Solution {
public:
int rob(vector<int>& nums) {
vector<int> dp(nums.size(), 0);
dp[0] = nums[0];
if(nums.size() < 2) return dp[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];
}
};
题目2:213. 打家劫舍 II - 力扣(LeetCode)
相比于1来说就是将环形分解成两种情况,考虑首元素 考虑尾元素
class Solution {
public:
int func(vector<int>& nums) {
vector<int> dp(nums.size(), 0);
dp[0] = nums[0];
if(nums.size() < 2) return dp[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<int>& nums) {
if(nums.size() < 2) return nums[0];
vector<int> v1(nums.size() -1, 0);
vector<int> v2(nums.size() -1, 0);
for(int i = 0;i < nums.size() - 1;i++) {
v1[i] = nums[i];
v2[i] = nums[i + 1];
}
return max(func(v1), func(v2));
}
};
题目3:337. 打家劫舍 III - 力扣(LeetCode)
class Solution {
public:
vector<int> robroot(TreeNode* node) {
if(node == NULL) return {0, 0};
vector<int> leftdp = robroot(node->left);
vector<int> rightdp = robroot(node->right);
int val1 = node->val + leftdp[0] + rightdp[0];
int val2 = max(leftdp[0], leftdp[1]) + max(rightdp[0], rightdp[1]);
return {val2, val1};
}
int rob(TreeNode* root) {
return max(robroot(root)[0], robroot(root)[1]);
}
};