编辑距离及编辑距离算法
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#include<iostream> #include<string> #include<algorithm> using namespace std; const int MAX = 1001; int MaxLen[MAX][MAX]; int maxLen(string str1, string str2) { int len1 = str1.length();//行 int len2 = str2.length();//列 for (int i = 0; i < len1; i++) MaxLen[i][0] = 0; for (int j = 0; j < len2; j++) MaxLen[0][j] = 0; for (int i = 1; i <= len1; i++) { for (int j = 1; j <= len2; j++) { if (str1[i - 1] == str2[j - 1]) MaxLen[i][j] = MaxLen[i - 1][j - 1] + 1; else { int temp = max(MaxLen[i - 1][j], MaxLen[i][j - 1]); MaxLen[i][j]=max(temp,MaxLen[i-1][j-1]); } } } return MaxLen[len1][len2]; } int main() { string str; int count = 0; while (cin >> str) { int len = str.size(); if (len == 1) { cout << 1 << endl; continue; } string revs = str; reverse(revs.begin(), revs.end()); int max_len = maxLen(str, revs); cout << len - max_len << endl; } return 0; }
参考:http://www.cnblogs.com/biyeymyhjob/archive/2012/09/28/2707343.html
这是很经典的动态规划问题。注意其中二维动态数组内存的分配和释放。
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int edit(const string str1, const string str2) { int m = str1.size(); int n = str2.size(); //定义一个m*n的二维数组 int **ptr = new int*[m+1]; for (int i = 0; i < m + 1; i++) ptr[i] = new int[n + 1]; //初始化 for (int i = 0; i < m + 1; i++) ptr[i][0] = i; for (int j = 0; j < n + 1; j++) ptr[0][j] = j; for (int i = 1; i < m + 1; i++) { for (int j = 1; j < n + 1; j++) { int d; int temp = min(ptr[i - 1][j] + 1, ptr[i][j - 1] + 1); if (str1[i - 1] == str2[j - 1]) d = 0; else d = 1; ptr[i][j] = min(temp, ptr[i - 1][j - 1] + d); } } int dis = ptr[m][n]; //注意释放内存 for (int i = 0; i < m + 1; i++) { delete[] ptr[i]; ptr[i] = nullptr; } delete[] ptr; ptr = nullptr; return dis; } int main() { string str1; string str2; cin >> str1; cin >> str2; edit(str1, str2); return 0; }
2017腾讯实习题,求最长公共子串的:http://www.nowcoder.com/test/question/done?tid=4822903&qid=44802
同样用DP求子符串与其反串的最长公共子串的长度,然后用总长减去公共子串的长度即可。