?

#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; } View Code

?

?

參考：http://www.cnblogs.com/biyeymyhjob/archive/2012/09/28/2707343.html

? ? ? ? ?這是很經典的動態規劃問題。注意其中二維動態數組內存的分配和釋放。

? ? ? ? ?

int edit(const string str1, const string str2) {int m = str1.size();int n = str2.size();//定義一個m*n的二維數組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;elsed = 1;ptr[i][j] = min(temp, ptr[i - 1][j - 1] + d);}}int dis = ptr[m][n];//注意釋放內存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; } View Code

?

?

2017騰訊實習題，求最長公共子串的：http://www.nowcoder.com/test/question/done?tid=4822903&qid=44802

?

同樣用DP求子符串與其反串的最長公共子串的長度，然后用總長減去公共子串的長度即可。

?

轉載于:https://www.cnblogs.com/573177885qq/p/5832328.html

總結

以上是生活随笔為你收集整理的编辑距离及编辑距离算法的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。