算法（DP）：最长公共子序列问题

最长公共子序列问题大体分为两种方法：

1.蛮力搜索：列举A所有的2^n个子序列，对于每一个子序列在θ(m)时间内来确定它是否也是B的子序列，复杂性是θ(m*2^n)，是指数复杂性的。

2.动态规划：最优解能够在θ(m*n)时间和θ(min{m,n})空间内得到。

public class LCS {
public static void main(String[] args) {
String[] a = {"x","y","x","x","z","x","y","z","x","y"};
String[] b = {"z","x","z","y","y","z","x","x","y","x","x","z"};
Lcss lcss = new Lcss();
lcss.longestcommonsubsequence(a, b);
}
}
class Lcss{
public void longestcommonsubsequence(String[] a, String[] b) {
int n = a.length + 1;
int m = b.length + 1;
int L[][] = new int[n][m];
for(int i = 0;i<n;i++){
L[i][0] = 0;
}
for(int j = 0;j<m;j++){
L[0][j] = 0;
}
for(int i = 1;i<n;i++){
for(int j = 1;j<m;j++){
if(a[i-1] == b[j-1]){                       //注意数组a，b是从0下标开始记录数据的，而存放最长公共子序列长度的数组L是从下标1开始计数的。
L[i][j] = L[i-1][j-1] + 1;
}else{
L[i][j] = max(L[i-1][j],L[i][j-1]);
}
}
}
myPrint(L);
}
private void myPrint(int[][] l) {
int n = l.length;
int m = l[1].length;
for(int i = 0;i<n;i++){
for(int j = 0;j<m;j++){
System.out.print(l[i][j]);
}
System.out.println("");
}
}
private int max(int i, int j) {
if(i>=j) {
return i;
}else{
return j;
}
}
}