最大子串和，最长连续子串，最长子序列（不连续）

 

package arithmetic;import java.util.Arrays;public class SumOfSubSequence {/* * 求组数的最大子串和(连续) */public static int maxSum(int data[], int n) {int sum, max;sum = max = data[0];for (int i = 0; i < n; i++) {if (sum < 0)sum = data[i];elsesum += data[i];if (sum > max)max = sum;}return max;}/* * 求最长公共子串(连续);只返回长度，不返回匹配的序列 */public static int[] getMatchNum(int a[], int b[])// 只返回长度，不返回子序列{int max = 0, c[] = new int[b.length];// 用一个数组就可以表示它的左上角的数是多少int pos = 0;for (int i = 0; i < a.length; i++) {for (int j = b.length-1; j >=0; j--) { //这里为什么是要从后向前比较？和那个用一维数组求0-1背包一样，都要从后向前比较if (a[i] == b[j] && j == 0)c[j] = 1;else if (a[i] == b[j])c[j] = c[j - 1] + 1;elsec[j] = 0;if (c[j] > max) {max = c[j]; // 记录最多匹配连续的个数pos = j;}}//System.out.println(Arrays.toString(c));}int result[] = new int[max];for (int i = pos - max + 1; i <= pos; i++)result[i - pos + max - 1] = b[i];return result;}/* * 重载上面的方法 * */public static String getMatch(String a,String b){int c[] = new int[b.length()];StringBuffer result = new StringBuffer();int max = 0,pos = 0;for(int i=0;i<a.length();i++)for(int j=b.length()-1;j>=0;j--){if(a.charAt(i)==b.charAt(j)&& j==0)c[j] = 1;else if(a.charAt(i)==b.charAt(j))c[j]=c[j-1]+1;elsec[j] = 0;if(max<c[j]){max = c[j];pos = j;}}for(int i=pos-max+1;i<=pos;i++)result.append(b.charAt(i));return result.toString();}/* * 最长公共子序列（不连续） * 公式：c[i][j] = |-0 (i=0||j=0) * |-c[i-1]c[j-1](a[i]==b[j]) * |-max(c[i][j-1],c[i-1][j]) * */public static int[][] getPath(String a,String b){int c [][] = new int[a.length()+1][b.length()+1];int path[][] = new int[a.length()+1][b.length()+1]; //定义路线    左上角：0 左边：－1 上边：1//  因为都是从1 下标开始，对String 进行处理a = "_"+a;b = "_"+b;for(int i=1;i<a.length();i++)for(int j=1;j<b.length();j++){if(a.charAt(i)==b.charAt(j)){c[i][j] = c[i-1][j-1]+1;path[i][j] = 1;}elseif(c[i-1][j]>=c[i][j-1])// 上边>左边{c[i][j] = c[i-1][j];// 上边path[i][j] = 2;}else{c[i][j] = c[i][j-1];path[i][j] = 3;// 左边}}return path;}public static void getCloseMath(int i,int j,String a,int[][] path){if(i==0||j==0)return;if(path[i][j]==1){getCloseMath(i-1,j-1,a,path);System.out.print(a.charAt(i-1));}elseif(path[i][j]==2)getCloseMath(i-1,j,a,path);elsegetCloseMath(i,j-1,a,path);}public static void main(String args[]) {int a[] = {0,1, 2, 3, 4, 0, 5, 6, 7, 8 };// 作为行int b[] = {0,1, 2, 3, 4, 9, 5, 6, 7, 8 };// 作为例System.out.println(Arrays.toString(getMatchNum(a, b)));String as = "123";String bs = "43678";System.out.println(getMatch(as,bs));String rs = "ABCBDAB";String cs = "BDCABA";int c[][] = getPath(rs,cs);getCloseMath(c.length-1,c[0].length-1,rs,c);}}