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1.打家劫舍

198. 打家劫舍 - 力扣（LeetCode）

1.1一维数组

class Solution
{
public:int rob(vector<int> &nums){int n = nums.size();if (n == 0)return 0;// 房子编号0~n-1// dp[k] 从0偷到k获得的最大金额vector<int> dp(n, 0);dp[0] = nums[0];if (n == 1)return dp[0];dp[1] = max(nums[0], nums[1]);if (n == 2)return dp[1];for (int k = 2; k <= n - 1; k++){dp[k] = max(dp[k - 1], nums[k] + dp[k - 2]);}return dp[n - 1];}
};

1.2三变量法

class Solution
{
public:int rob(vector<int> &nums){int n = nums.size();if (n == 0)return 0;int prv = nums[0]; // dp[0]if (n == 1)return prv;int cur = max(nums[0], nums[1]); // dp[1]if (n == 2)return cur;// 房子编号0~n-1// dp[k] 从0偷到k获得的最大金额for (int i = 2; i < n; i++){// dp[k] = max(dp[k - 1], nums[k] + dp[k - 2]);int tmp = max(cur, nums[i] + prv);prv = cur;cur = tmp;}return cur;}
};

1.3双数组法

class Solution
{
public:int massage(vector<int> &nums){int n = nums.size();if (n == 0)return 0;// f[i] 走到i时 偷nums[i]// g[i] 走到i时 不偷nums[i]vector<int> f(n);vector<int> g(n); // auto g = f;f[0] = nums[0];for (int i = 1; i < n; i++){f[i] = nums[i] + g[i - 1];g[i] = max(f[i - 1], g[i - 1]);}return max(f[n - 1], g[n - 1]);}
};

2.打家劫舍2

打家劫舍2

2.1双数组法

class Solution {
public:int rob(vector<int>& nums) {int n = nums.size();return max(nums[0] + rob1(nums, 2, n - 2), rob1(nums, 1, n - 1));}int rob1(vector<int>& nums, int left, int right) {if (left > right)return 0;int n = nums.size();vector<int> f(n);auto g = f;f[left] = nums[left];for (int i = left + 1; i <= right; i++) {f[i] = g[i - 1] + nums[i];g[i] = max(f[i - 1], g[i - 1]);}return max(f[right], g[right]);}
};

2.2 三变量法

class Solution
{
public:int robRange(vector<int> &nums, int start, int end){int n = end - start + 1;if (n == 0)return 0;int prv = nums[start]; // dp[k-2]if (n == 1)return prv;int cur = max(nums[start], nums[start + 1]); // dp[k-1]if (n == 2)return cur;for (int i = start + 2; i <= end; i++){// dp[k] = max(dp[k - 1], nums[k - 1] + dp[k - 2]);int tmp = max(cur, nums[i] + prv);prv = cur;cur = tmp;}return cur;}int rob(vector<int> &nums){int n = nums.size();if (n == 1)return nums[0];else if (n == 2)return max(nums[0], nums[1]);else if (n == 3)return max(max(nums[0], nums[1]), nums[2]);// 偷nums[0]不能偷nums[1]和nums[n-1] 能偷[2, n - 2]// 不偷nums[0] 能偷[1, n - 1]return max(nums[0] + robRange(nums, 2, n - 2), robRange(nums, 1, n - 1));}
};

3.打家劫舍3

3.1动态规划

class Solution
{
public:unordered_map<TreeNode *, int> is, no;void dfs(TreeNode *node){if (node == nullptr)return;dfs(node->left);dfs(node->right);is[node] = node->val + no[node->left] + no[node->right];no[node] = max(is[node->left], no[node->left]) + max(is[node->right], no[node->right]);}int rob(TreeNode *root){dfs(root);return max(is[root], no[root]);}
};

3.2双变量法

struct SubtreeStatus
{int is;int no;
};class Solution
{
public:SubtreeStatus dfs(TreeNode *node){if (node == nullptr)return {0, 0};auto left = dfs(node->left);auto right = dfs(node->right);int selected = node->val + left.no + right.no;int notSelected = max(left.is, left.no) + max(right.is, right.no);return {selected, notSelected};}int rob(TreeNode *root){auto rootStatus = dfs(root);return max(rootStatus.is, rootStatus.no);}
};

4.删除相邻数字的最大分数

删除相邻数字的最大分数_牛客题霸_牛客网 (nowcoder.com)

1. 一个数组选了x可以得x分 但是值为x-1和x+1的数将会消失 直到数字全被消失或选择 问最高分数
2. 遍历数组arr 填充hash arr中的per在hash中作下标 表示 【谁 出现的总和】如4在arr中出现了2次 则hash[4]=8
3. 由此问题转变为 在hash中 我“偷”了某个下标i 获得了hash[i]分数 与i相邻的就不能偷了

4.1双状态数组

#include <iostream>
#include <vector>
using namespace std;int main()
{const int N = 1e4 + 10;int hash[N] = {0};int n = 0;cin >> n;int per = 0;int end = 0;for (int i = 0; i < n; i++){cin >> per;end = per > end ? per : end;hash[per] += per;}vector<int> f(end + 1, 0);vector<int> g(end + 1, 0);for (int i = 1; i <= end; i++){f[i] = g[i - 1] + hash[i];g[i] = max(f[i - 1], g[i - 1]);}cout << max(f[end], g[end]) << endl;return 0;
}

4.2一维数组

#include <iostream>
#include <vector>
using namespace std;int main()
{int n = 0;cin >> n;int per = 0;int end = 0;const int N = 1e4 + 10;int hash[N] = {0};for (int i = 0; i < n; i++){cin >> per;end = per > end ? per : end;hash[per] += per;}vector<int> dp(end + 1, 0);// 房子编号0~n-1// dp[k] 从0偷到k获得的最大金额dp[0] = hash[0];dp[1] = max(hash[0], hash[1]);for (int k = 2; k <= end; k++){dp[k] = max(dp[k - 1], hash[k] + dp[k - 2]);}cout << dp[end] << endl;return 0;
}

4.3三变量法

#include <iostream>
#include <vector>
using namespace std;int main()
{const int N = 1e4 + 10;int hash[N] = {0};int n = 0;cin >> n;int per = 0;int end = 0;for (int i = 0; i < n; i++){cin >> per;end = per > end ? per : end;hash[per] += per;}// 房子编号0~n-1// dp[k] 从0偷到k获得的最大金额int prv = hash[0];               // dp[0]int cur = max(hash[0], hash[1]); // dp[1]for (int i = 2; i <= end; i++){// dp[k] = max(dp[k - 1], hash[k] + dp[k - 2]);int tmp = max(cur, hash[i] + prv);prv = cur;cur = tmp;}cout << cur << endl;return 0;
}