先来一个简单dp练习

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

我们只需要考虑前面的就行，后面的就交给后面的元素考虑

再来一个类似的题目

这个题目咋一看还没有什么思路，但是我们可以转化为打家劫舍

class Solution {
public:int deleteAndEarn(vector<int>& nums) {int mmax = 0;for (auto u : nums) {mmax = max(mmax, u);}vector<int> dp(mmax + 1);for (auto u : nums) {dp[u] += u;}return fun(dp);}int fun(vector<int> nums) {int n = nums.size();vector<int> dp(n + 1);int ans = 0;ans = dp[0] = nums[0];if (n == 1) return ans;dp[1] = max(nums[0], nums[1]);ans = max(ans, dp[1]);if (n == 2) return ans;for (int i = 2; i <= n; i++) {dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);ans = max(dp[i], ans);}return ans;}
};

可是如果我们数组里面的数字很大怎么办，我们不能把我们的数组也开的特别大

class Solution {
public:long long maximumTotalDamage(vector<int>& power) {unordered_map<int, int> cnt;for (auto u : power) {cnt[u]++;}vector<pair<int, int>> a(cnt.begin(), cnt.end()); // 拷贝过来，此处很重要// 需要排序吗ranges::sort(a);int n = a.size();vector<long long> dp(n + 1);for (int i = 0, j = 0; i < n; i++) {auto& [x, c] = a[i];while (a[j].first < x - 2) {j++;}dp[i + 1] = max(dp[i], dp[j] + (long long)x * c);}return dp[n];}};