第6章 第6节 动态规划
● 请你手写代码：最长公共连续子序列
参考回答：
int substr(string & str1, string &str2){int len1 = str1.length();int len2 = str2.length();vector<vector<int>>dp(len1,vector<int>(len2,0));for (int i = 0; i < len1; i++){dp[i][0] = str1[i]==str1[0]?1:0;}for (int j = 0; j <= len2; j++){dp[0][j] = str1[0]==str2[j]?1:0;}for (int i = 1; i < len1; i++){for (int j = 1; j < len2; j++){if (str1[i] == str2[j]){dp[i][j] = dp[i - 1][j - 1]+1;}}}int longest = 0;int longest_index = 0;for (int i = 0; i < len1; i++){for (int j = 0; j < len2; j++){if (longest < dp[i][j]){longest = dp[i][j];longest_index = i;}}}
//字符串为从第i个开始往前数longest个
for (int i = longest_index-longest+1; i <=longest_index; i++){cout << str1[i] << endl;}return longest;}
● 手写代码：求一个字符串最长回文子串
参考回答：
int LongestPalindromicSubstring(string & a){int len = a.length();vector<vector<int>>dp(len, vector<int>(len, 0));for (int i = 0; i < len; i++){dp[i][i] = 1;}int max_len = 1;int start_index = 0;for (int i= len - 2; i >= 0; i--){for (int j = i + 1; j < len; j++){if (a[i] == a[j]){if (j - i == 1){dp[i][j] = 2;}else{if (j - i > 1){dp[i][j] = dp[i + 1][j - 1] + 2;}}if (max_len < dp[i][j]){max_len = dp[i][j];start_index = i;}}else{dp[i][j] = 0;}}}cout << "max len is " << max_len << endl;cout << "star index is" << start_index << endl;return max_len;}
● 手写代码：查找最长回文子串
参考回答：
int LongestPalindromicSubstring(string & a){int len = a.length();vector<vector<int>>dp(len, vector<int>(len, 0));for (int i = 0; i < len; i++){dp[i][i] = 1;}int max_len = 1;int start_index = 0;for (int i= len - 2; i >= 0; i--){for (int j = i + 1; j < len; j++){if (a[i] == a[j]){if (j - i == 1){dp[i][j] = 2;}else{if (j - i > 1){dp[i][j] = dp[i + 1][j - 1] + 2;}}if (max_len < dp[i][j]){max_len = dp[i][j];start_index = i;}}else{dp[i][j] = 0;}}}cout << "max len is " << max_len << endl;cout << "star index is" << start_index << endl;return max_len;}