最长公共子序列&最长公共子串

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#include<iostream>#include<string>#include<algorithm>#include<vector>using namespace std;/************************************************求最长公共子序列长度*************************************************/int LCS(string str1, string str2) {int len1 = str1.length();int len2 = str2.length();if (len1 == 0 || len2 == 0) return 0;vector<vector<int>> dp(len1, vector<int>(len2, 0));dp[0][0] = str1[0] == str2[0] ? 1 : 0;for (int i = 1; i < len1; i++) dp[i][0] = str1[i] == str2[0] ? 1 : 0; for (int i = 1; i < len2; i++) dp[0][i] = str1[0] == str2[i] ? 1 : 0;for (int i = 1; i < len1; i++) {for (int j = 1; j < len2; j++) {if (str1[i] == str2[j]) {dp[i][j] = 1 + dp[i - 1][j - 1];}else {dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);}}}return dp[len1 - 1][len2 - 1];}/************************************************求最长公共子串 *************************************************//*string LCSS(string str1, string str2) {int len1 = str1.length();int len2 = str2.length();if (len1 == 0 || len2 == 0) return "";vector<vector<int>> dp(len1, vector<int>(len2, 0));dp[0][0] = str1[0] == str2[0] ? 1 : 0;for (int i = 1; i < len1; i++) dp[i][0] = str1[i] == str2[0] ? 1 : 0;for (int i = 1; i < len2; i++) dp[0][i] = str1[0] == str2[i] ? 1 : 0;for (int i = 1; i < len1; i++) {for (int j = 1; j < len2; j++) {if (str1[i] == str2[j]) {dp[i][j] = 1 + dp[i - 1][j - 1];}}}int end = 0;int max_len = 0;for (int i = 0; i < len1; i++) {for (int j = 0; j < len2; j++) {if (max_len < dp[i][j]) {max_len = dp[i][j];end = i;}}}return str1.substr(end - max_len + 1, max_len);}*///版本二：O(1)的空间复杂度string LCSS(string str1, string str2) {int len1 = str1.length();int len2 = str2.length();if (len1 == 0 || len2 == 0) return "";int max_len = 0;int end = 0;int row = 0, col = len2 - 1, len;while (row < len1) {int i = row, j = col;len = 0;while (i < len1&&j < len2) {if (str1[i] != str2[j]) len = 0;else len++;if (len > max_len) {end = i;max_len = len;}i++; j++;}if (col > 0) col--;else row++;}return str1.substr(end - max_len + 1, max_len);}int main(void) {int a[] = { 1, -1, 2, -1, 4, -5, 6, -7 };cout << LCSS("1D23D4B56", "B1D23CA45B6A") << endl;return 0;}
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