力扣hot100——动态规划 多维动态规划

前言：题太多了TAT，只贴了部分我觉得比较好的题 

32. 最长有效括号

class Solution {
public:int longestValidParentheses(string s) {int n = s.size();s = " " + s;vector<int> dp(n + 1, 0);int ans = 0;for (int i = 2; i <= n; i++) {if (s[i] == '(') dp[i] = 0;else {if (s[i - 1] == '(') dp[i] = dp[i - 2] + 2;else {int pre = i - 2 - dp[i - 1];if (s[pre + 1] == '(') dp[i] = dp[i - 1] + 2 + (pre > 0 ? dp[pre] : 0);}}ans = max(ans, dp[i]);}return ans;}
};

分类讨论：
si == '(' : 肯定dpi = 0
si == ‘)’: 看si - 1的是不是'('， 如果不是再往前找...
注意dp[pre]会连接起来

152. 乘积最大子数组

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

分类大讨论，维护以i为结尾的乘积最大值和最小值即可

72. 编辑距离

class Solution {
public:int minDistance(string s1, string s2) {int n = s1.size(), m = s2.size();if (!n) {return m;}if (!m) {return n;}s1 = " " + s1;s2 = " " + s2;vector<vector<int>> dp(n + 1, vector<int>(m + 1, 1e5));dp[0][0] = 0;for (int i = 1; i <= n; i++) dp[i][0] = i;for (int i = 1; i <= m; i++) dp[0][i] = i;for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {int t = min({ dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1] }) + 1;dp[i][j] = min(dp[i][j], t);if (s1[i] == s2[j]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);}}return dp[n][m];}
};

5. 最长回文子串

class Solution {
public:string longestPalindrome(string s) {string str = "";str.push_back('$');str.push_back('#');for (auto ch : s) {str.push_back(ch);str.push_back('#');}int n = str.size() - 1;vector<int> d(n + 10, 0);d[1] = 1;for (int i = 2, l, r = 1; i <= n; i++) {if (i <= r) d[i] = min(d[l + r - i], r - i + 1);while (str[i - d[i]] == str[i + d[i]]) d[i]++;if (i + d[i] - 1 > r) {l = i - d[i] + 1;r = i + d[i] - 1;}}int ans = 0;int p = 0;string ss = "";for (int i = 1; i <= n; i++) {if (ans < d[i] - 1) {ans = d[i] - 1;p = i - d[i] + 1;}}for (int i = p; ans > 0 ; i++) {if (str[i] != '#') {ss += str[i];ans--;}}return ss;}
};