（1）理论部分：
（2）习题：
最长公共子串：
1 package month7.dp; 2 3 //https://www.nowcoder.com/questionTerminal/181a1a71c7574266ad07f9739f791506 4 public class 最长公共子串 { 5 public static void main(String[] args) { 6 最长公共子串 s = new 最长公共子串(); 7 String str1 = "ABABCGGG"; 8 String str2 = "BABCAH"; 9 System.out.println(s.Longest_Common_Substring(str1, str2)); 10 } 11 //将二维数组table[i][j]用来记录具有这样特点的子串——结尾同时也为为串x1x2⋯xi与y1y2⋯yj的结尾——的长度 12 public int Longest_Common_Substring(String str1, String str2) { 13 int len1 = str1.length(); 14 int len2 = str2.length(); 15 int result = 0; 16 int num = 0; 17 int[][] table = new int[len1 + 1][len2 + 1]; 18 19 for (int i = 0; i <= len1; i++) { 20 table[i][0] = 0; 21 } 22 for (int j = 0; j <= len2; j++) { 23 table[0][j] = 0; 24 } 25 for (int i = 1; i <= len1; i++) { 26 for (int j = 1; j <= len2; j++) { 27 char c1 = str1.charAt(i-1); 28 char c2 = str2.charAt(j-1); 29 if (c1 != c2) { 30 table[i][j] = 0; 31 } else { // c1 == c2 32 table[i][j] = table[i - 1][j - 1] + 1; 33 if (result < table[i][j]) { 34 result = table[i][j]; 35 num = i; 36 System.out.println("-----" + num + "--"); 37 38 } 39 } 40 } 41 } 42 System.out.println(str1.substring(num - result , num )); 43 return result; 44 } 45 46 public int Longest_Common_Substring2(String str1, String str2) { 47 int len1 = str1.length(); 48 int len2 = str2.length(); 49 int result = 0; 50 int num = 0; 51 int[][] table = new int[len1 + 1][len2 + 1]; 52 for (int i = 0; i <= len1; i++) { 53 for (int j = 0; j <= len2; j++) { 54 if (i == 0 || j == 0) { 55 table[i][j] = 0; 56 } else if (str1.charAt(i - 1) != str2.charAt(j - 1)) { 57 table[i][j] = 0; 58 } else { // c1 == c2 59 table[i][j] = table[i - 1][j - 1] + 1; 60 if (result < table[i][j]) { 61 result = table[i][j]; 62 num = i; 63 System.out.println("-----" + num + "--"); 64 65 } 66 } 67 } 68 } 69 System.out.println(str1.substring(num - result , num )); 70 return result; 71 } 72 73 }
最长公共子序列：
1 package month7.dp; 2 3 public class 最长公共子序列 { 4 int[][] b = new int[10][10]; 5 public static void main(String[] args) { 6 最长公共子序列 s = new 最长公共子序列(); 7 // String str1 ="ABCD"; 8 // String str2 = "EACB"; 9 String str1 ="AGGTHAB"; 10 String str2 = "GXTXAYB"; 11 System.out.println(s.Longest_Common_Subsequence(str1, str2)); 12 s.print_lsc(str1.length(), str2.length(), str1); 13 14 } 15 //用二维数组c[i][j]记录串x1x2⋯xi与y1y2⋯yj的LCS长度 16 public int Longest_Common_Subsequence(String str1,String str2){ 17 int len1 = str1.length(); 18 int len2 = str2.length(); 19 int[][] table = new int[str1.length()+1][str2.length()+1]; 20 // int[][] b = new int[str1.length()+1][str2.length()+1]; 21 for(int i=0;i<=len1;i++){ 22 for(int j=0;j<=len2;j++){ 23 if(i==0 || j == 0){ 24 table[i][j] = 0; 25 }else if(str1.charAt(i-1) != str2.charAt(j-1)){ 26 if(table[i-1][j] > table[i][j-1]){ 27 table[i][j] = table[i-1][j]; 28 b[i][j] = 1; 29 }else{ 30 table[i][j] = table[i][j-1]; 31 b[i][j] = 2; 32 } 33 // table[i][j] = Math.max(table[i-1][j], table[i][j-1]); 34 }else{ // Xi = Yj 35 table[i][j] = table[i-1][j-1] +1; 36 b[i][j] = 3; 37 } 38 } 39 } 40 int commomLength = table[len1][len2]; 41 return commomLength; 42 } 43 public void print_lsc(int i,int j,String str1){ 44 if(i ==0 || j == 0) 45 return ; 46 if(b[i][j] == 3){ 47 print_lsc(i-1, j-1, str1); 48 System.out.print(str1.charAt(i-1)); 49 }else if(b[i][j] == 2){ 50 print_lsc(i, j-1, str1); 51 }else if(b[i][j] == 1){ 52 print_lsc(i-1, j, str1); 53 } 54 } 55 }
最短编辑距离
1 package month7.dp; 2 3 //https://www.cnblogs.com/masterlibin/p/5785092.html 4 public class 最短编辑距离 { 5 public static void main(String[] args) { 6 最短编辑距离 s = new 最短编辑距离(); 7 8 } 9 10 public int editionDistance(String word1, String word2) { 11 if (word1 == null || "".equals(word1)) { 12 return word2.length(); 13 } 14 if (word2 == null || "".equals(word1)) { 15 return word1.length(); 16 } 17 int len1 = word1.length(); 18 int len2 = word2.length(); 19 int[][] table = new int[len1 + 1][len2 + 1]; 20 for (int i = 0; i <= len1; i++) { 21 table[i][0] = i; 22 } 23 for (int j = 0; j <= len2; j++) { 24 table[0][j] = j; 25 } 26 for (int i = 1; i <= len1; i++) { 27 for (int j = 1; j <= len2; j++) { 28 char c1 = word1.charAt(i - 1); 29 char c2 = word2.charAt(j - 1); 30 if (c1 == c2) { 31 table[i][j] = table[i - 1][j - 1]; 32 } else { 33 int add = table[i][j - 1] + 1; 34 int delate = table[i - 1][j] + 1; 35 int edition = table[i - 1][j - 1] + 1; 36 table[i][j] = Math.min(add, Math.min(delate, edition)); 37 } 38 } 39 } 40 return table[len1][len2]; 41 } 42 43 }