【力扣系列题目】最后一块石头的重量分割回文串验证回文串等差数列划分{最大堆背包动态规划}

文章目录

七、最后一块石头的重量
- 最后一块石头的重量【堆】
- [最后一块石头的重量 II](https://leetcode.cn/problems/last-stone-weight-ii/)【背包】
八、分割回文串
- 分割回文串【分割子串方案数量】
- [分割回文串 II](https://leetcode.cn/problems/omKAoA/)【最少分割次数】
- [分割回文串 III](https://leetcode.cn/problems/palindrome-partitioning-iii/)【分割成k个+可修改】
- [分割回文串 IV](https://leetcode.cn/problems/palindrome-partitioning-iv/)【是否能发成三个】
九、验证回文串
- 验证回文串
- [验证回文串 II](https://leetcode.cn/problems/RQku0D/)【最多删除一个】
- [验证回文串 III](https://leetcode.cn/problems/valid-palindrome-iii/)【可删除k个】
- [验证回文串 IV](https://leetcode.cn/problems/valid-palindrome-iv/)【可删除一或两个】
十、等差数列划分
- 前言
- 等差数列划分【等差数组的子数组个数】
- [等差数列划分 II - 子序列](https://leetcode.cn/problems/arithmetic-slices-ii-subsequence/)
十一、demo

七、最后一块石头的重量

最后一块石头的重量【堆】

class Solution {
public:int lastStoneWeight(vector<int>& stones) {priority_queue<int> q;for (int s : stones) {q.push(s);}while (q.size() > 1) {int a = q.top();q.pop();int b = q.top();q.pop();if (a > b) {q.push(a - b);}}return q.empty() ? 0 : q.top();}
};

最后一块石头的重量 II【背包】

class Solution {
public:int lastStoneWeightII(vector<int>& stones) {int sum = 0;for (auto x : stones)sum += x;int n = stones.size(), m = sum / 2;vector<vector<int>> dp(n + 1, vector<int>(m + 1));for (int i = 1; i <= n; i++) {for (int j = 0; j <= m; j++) {dp[i][j] = dp[i - 1][j];if (j >= stones[i - 1])dp[i][j] = max(dp[i][j], dp[i - 1][j - stones[i - 1]] +stones[i - 1]);}}return sum - 2 * dp[n][m];}
};

八、分割回文串

分割回文串【分割子串方案数量】

class Solution {
private:vector<vector<int>> f;vector<vector<string>> ans;vector<string> path;int n;void dfs(const string& s, int i) {if (i == n) {ans.push_back(path);return;}for (int j = i; j < n; ++j) {if (isPalindrome(s, i, j) == 1) {path.push_back(s.substr(i, j - i + 1));dfs(s, j + 1);path.pop_back();}}}// 0未搜索 1回文串 -1不是回文串int isPalindrome(const string& s, int i, int j) {if (f[i][j] != 0)return f[i][j];if (i > j || i == j || (i + 1 == j && s[i] == s[j]))return f[i][j] = 1;f[i][j] = (s[i] == s[j] ? isPalindrome(s, i + 1, j - 1) : -1);return f[i][j];}public:vector<vector<string>> partition(string s) {n = s.size();f.assign(n, vector<int>(n));dfs(s, 0);return ans;}
};

分割回文串 II【最少分割次数】

class Solution
{void init(const string &s, vector<vector<bool>> &isPal){int n = s.size();for (int i = n - 1; i >= 0; i--){for (int j = i; j < n; j++){if (i == j || (i + 1 == j && s[i] == s[j]))isPal[i][j] = true;else if (s[i] == s[j])isPal[i][j] = isPal[i + 1][j - 1];}}}public:int minCut(string s){int n = s.size();vector<vector<bool>> isPal(n, vector<bool>(n,false));init(s, isPal);vector<int> dp(n, INT_MAX);for (int i = 0; i < n; i++){if (isPal[0][i])dp[i] = 0;else{ for (int j = 1; j <= i; j++)if (isPal[j][i])dp[i] = min(dp[i], dp[j - 1] + 1);}}return dp[n - 1];}
};

分割回文串 III【分割成k个+可修改】

class Solution {int cost2(string& s, int l, int r) {int ret = 0;for (int i = l, j = r; i < j; ++i, --j) {if (s[i] != s[j])++ret;}return ret;}void init(string& s, vector<vector<int>>& cost) {int n = s.size();// span：区间长度; i：区间起点; j：区间终点for (int span = 2; span <= n; ++span) {for (int i = 0; i <= n - span; ++i) {int j = i + span - 1;cost[i][j] = cost[i + 1][j - 1] + (s[i] == s[j] ? 0 : 1);}}}public:int palindromePartition(string& s, int k) {int n = s.size();vector<vector<int>> cost(n, vector<int>(n));init(s, cost);vector<vector<int>> f(n + 1, vector<int>(k + 1, INT_MAX));f[0][0] = 0;for (int i = 1; i <= n; ++i) {for (int j = 1; j <= min(k, i); ++j) {if (j == 1)// f[i][j] = cost(s, 0, i - 1);f[i][j] = cost[0][i - 1];else {for (int i0 = j - 1; i0 < i; ++i0) {f[i][j] =// min(f[i][j], f[i0][j - 1] + cost(s, i0, i - 1));min(f[i][j], f[i0][j - 1] + cost[i0][i - 1]);}}}}return f[n][k];}
};

分割回文串 IV【是否能发成三个】

class Solution {void init(const string& s, vector<vector<bool>>& isPal) {int n = s.size();for (int i = n - 1; i >= 0; i--) {for (int j = i; j < n; j++) {if (i == j || (i + 1 == j && s[i] == s[j]))isPal[i][j] = true;else if (s[i] == s[j])isPal[i][j] = isPal[i + 1][j - 1];}}}public:bool checkPartitioning(string s) {int n = s.size();vector<vector<bool>> dp(n, vector<bool>(n));init(s, dp);for (int i = 1; i < n - 1; i++)for (int j = i; j < n - 1; j++)if (dp[0][i - 1] && dp[i][j] && dp[j + 1][n - 1])return true;return false;}
};

九、验证回文串

验证回文串

class Solution {
public:bool isPalindrome(string s) {int n = s.size();int left = 0, right = n - 1;while (left < right) {while (left < right && !isalnum(s[left]))++left;while (left < right && !isalnum(s[right]))--right;if (left < right) {if (tolower(s[left]) != tolower(s[right]))return false;++left;--right;}}return true;}
};

验证回文串 II【最多删除一个】

class Solution {
public:bool checkPalindrome(const string& s, int low, int high) {int i = low, j = high;while (i < j) {if (s[i] != s[j])return false;++i;--j;}return true;}bool validPalindrome(string s) {int low = 0, high = s.size() - 1;while (low < high) {char c1 = s[low], c2 = s[high];if (c1 == c2) {++low;--high;} else {return checkPalindrome(s, low, high - 1) ||checkPalindrome(s, low + 1, high);}}return true;}
};

验证回文串 III【可删除k个】

class Solution {bool fun(string& s, int k) {int n = s.size();vector<vector<int>> f(n, vector<int>(n));for (int i = n - 1; i >= 0; i--) {for (int j = i + 1; j < n; j++) {if (i == j || (i + 1 == j && s[i] == s[j]))f[i][j] = 0;if (i + 1 == j && s[i] != s[j])f[i][j] = 1;if (s[i] == s[j])f[i][j] = f[i + 1][j - 1];elsef[i][j] = min(f[i][j - 1], f[i + 1][j]) + 1;}}return f[0][n - 1] <= k;}public:vector<vector<int>> memo;int dfs(string& s, int i, int j) {if (i == j)return 0;if (i + 1 == j)return s[i] == s[j] ? 0 : 1;if (memo[i][j] != -1)return memo[i][j];if (s[i] == s[j])return memo[i][j] = dfs(s, i + 1, j - 1);return memo[i][j] = 1 + min(dfs(s, i + 1, j), dfs(s, i, j - 1));}bool isValidPalindrome(string s, int k) {return fun(s, k);memo.resize(s.length(), vector<int>(s.length(), -1));return dfs(s, 0, s.length() - 1) <= k;}
};

验证回文串 IV【可删除一或两个】

class Solution {
public:bool makePalindrome(string s) {int need = 0;int l = 0, r = s.size() - 1;while (l < r) {if (s[l] != s[r]) need++;l++;r--;}return need <= 2;}
};

十、等差数列划分

前言

先看这道题

最长等差数列

按照固定i j后两位然后找k的存在的思想在本题中无法解决。因为该题的测试用例中，a可能存在多个，题目需要“最长”，当i前有多个a时，我们需要的是最近的a，所以初始化index数组时，不能将所有元素提前处理好，map中的键是唯一的，故，后来的a会覆盖之前的a，也就无法在枚举i时，特定的去i前找。

错误解法示例

class Solution
{
public:int longestArithSeqLength(vector<int> &arr){unordered_map<int, int> indices;int n = arr.size();for (int i = 0; i < n; i++)indices[arr[i]] = i;vector<vector<int>> dp(n, vector<int>(n,2));int ans = 2;for (int i = 2; i < n; i++){for (int j = i - 1; j >= 0; j--){// a b c  a=b-(c-b)=2b-c// k j iint k = -1;int x = 2 * arr[j] - arr[i];if (indices.count(x))k = indices[x];if (0 <= k && k < j)dp[j][i] = dp[k][j] + 1;ans = max(ans, dp[j][i]);}}return ans;}
};

非得按照之前的思路

class Solution {
public:int longestArithSeqLength(vector<int>& arr) {int n = arr.size();unordered_map<int, vector<int>> hash;for (int i = 0; i < n; i++)hash[arr[i]].push_back(i);vector<vector<int>> dp(n, vector<int>(n, 2));int ans = 2;// a b c  a=b-(c-b)=2b-c// k j ifor (int i = 2; i < n; i++) {for (int j = i - 1; j >= 0; j--) {int k = -1;int x = 2 * arr[j] - arr[i];if (hash.count(x)) {for (int k : hash[x]) {if (0 <= k && k < j)dp[j][i] = dp[k][j] + 1;}}ans = max(ans, dp[j][i]);}}return ans;}
};

class Solution
{
public:int longestArithSeqLength(vector<int> &nums){unordered_map<int, int> hash;hash[nums[0]] = 0;int n = nums.size();vector<vector<int>> dp(n, vector<int>(n, 2));int ret = 2;// a b c  a=b-(c-b)=2b-c// k j ifor (int j = 1; j < n; j++){for (int i = j + 1; i < n; i++){int a = 2 * nums[j] - nums[i];if (hash.count(a))dp[j][i] = dp[hash[a]][j] + 1;ret = max(ret, dp[j][i]);}hash[nums[j]] = j;}return ret;}
};

等差数列划分【等差数组的子数组个数】

class Solution {// a b c// a b c d && b c d
public:int numberOfArithmeticSlices(vector<int>& nums) {// dp[i] 表⽰必须「以 i 位置的元素为结尾」的等差数列有多少种。int n = nums.size();vector<int> dp(n);int sum = 0;for (int i = 2; i < n; i++) {dp[i] = nums[i] - nums[i - 1] == nums[i - 1] - nums[i - 2]? dp[i - 1] + 1: 0;sum += dp[i];}return sum;}
};

等差数列划分 II - 子序列

class Solution {
public:int numberOfArithmeticSlices(vector<int>& nums) {int n = nums.size();unordered_map<long long, vector<int>> hash;for (int i = 0; i < n; i++)hash[nums[i]].push_back(i);vector<vector<int>> dp(n, vector<int>(n));int sum = 0;for (int j = 2; j < n; j++){for (int i = 1; i < j; i++) {long long a = (long long)nums[i] * 2 - nums[j];if (hash.count(a))for (auto k : hash[a])if (k < i)dp[i][j] += dp[k][i] + 1;elsebreak;sum += dp[i][j];}}return sum;}
};