最长公共序列算法

LCS-LENGTH (X, Y)
 1. m ← length [X]										
 2. n ← length [Y]
 3. for i ← 1 to m
 4. do c [i, 0] ← 0
 5. for j ← 0 to m
 6. do c [0, j] ← 0
 7. for i ← 1 to m
 8. do for j ← 1 to n
 9. do if xi= yj	
 10. then c [i, j] ← c [i-1, j-1] + 1	
 11. b [i, j] ← "↖"
 12. else if c[i-1, j] ≥ c[i, j-1]
 13. then c [i, j] ← c [i-1, j]
 14. b [i, j] ← "↑"
 15. else c [i, j] ← c [i, j-1]
 16. b [i, j] ← "← "
 17. return c and b.

最长公共序列的示例

示例：给定两个序列X [1 … m]和Y [1 ….. n]。找到两者的最长共同子序列。

here X = (A, B, C, B, D, A, B) and Y = (B, D, C, A, B, A)
     m = length [X] and n = length [Y]
     m = 7 and n = 6
Here x1= x [1] = A   y1= y [1] = B
     x2= B  y2= D	
     x3= C  y3= C
     x4= B  y4= A
     x5= D  y5= B
     x6= A  y6= A
     x7= B
Now fill the values of c [i, j] in m x n table
Initially, for i=1 to 7 c [i, 0] = 0
          For j = 0 to 6 c [0, j] = 0

那是：

Now for i=1 and j = 1
	x1 and y1 we get x1 ≠ y1 i.e. A ≠ B
And 	c [i-1, j] = c [0, 1] = 0
	c [i, j-1] = c [1, 0 ] = 0
That is, c [i-1, j]= c [i, j-1] so c [1, 1] = 0 and b [1, 1] = ' ↑  '

Now for i=1 and j = 2
x1 and y2 we get x1 ≠ y2 i.e. A ≠ D
	c [i-1, j] = c [0, 2] = 0
	c [i, j-1] = c [1, 1 ] = 0
That is, c [i-1, j]= c [i, j-1] and c [1, 2] = 0 b [1, 2] = '  ↑  '

Now for i=1 and j = 3
	x1 and y3 we get x1 ≠ y3 i.e. A ≠ C
	c [i-1, j] = c [0, 3] = 0
	c [i, j-1] = c [1, 2 ] = 0
so c [1, 3] = 0     b [1, 3] = ' ↑ '

Now for i=1 and j = 4
	x1 and y4 we get. x1=y4 i.e A = A 
	 c [1, 4] = c [1-1, 4-1] + 1
		   = c [0, 3] + 1
 		   = 0 + 1 = 1
	c [1, 4] = 1
	b [1, 4] = '  ↖  '

Now for i=1 and j = 5
           x1 and y5  we get x1 ≠ y5
           c [i-1, j] = c [0, 5] = 0
	c [i, j-1] = c [1, 4 ] = 1
Thus c [i, j-1] >  c [i-1, j] i.e. c [1, 5] = c [i, j-1] = 1. So b [1, 5] = '←'

Now for i=1 and j = 6
           x1 and y6   we get x1=y6
                     c [1, 6] = c [1-1, 6-1] + 1
                              = c [0, 5] + 1 = 0 + 1 = 1
		   c [1, 6] = 1
		   b [1, 6] = '  ↖  '

Now for i=2 and j = 1
 We get x2 and y1 B = B i.e.  x2= y1
             c [2, 1] = c [2-1, 1-1] + 1
                     = c [1, 0] + 1
                     = 0 + 1 = 1   
             c [2, 1] = 1 and b [2, 1] = ' ↖ '
Similarly, we fill the all values of c [i, j] and we get

步骤4：构建LCS：初始调用为PRINT-LCS（b, X, X.length, Y.length）

PRINT-LCS (b, x, i, j)
 1. if i=0 or j=0
 2. then return
 3. if b [i, j] = ' ↖ '
 4. then PRINT-LCS (b, x, i-1, j-1)
 5. print x_i
 6. else if b [i, j] = '  ↑  '
 7. then PRINT-LCS (b, X, i-1, j)
 8. else PRINT-LCS (b, X, i, j-1)

示例：确定（1, 0, 0, 1, 0, 1, 0, 1）和（0, 1, 0, 1, 1, 0, 1, 1, 0）的LCS。

解：让X =（1, 0, 0, 1, 0, 1, 0, 1）和Y =（0, 1, 0, 1, 1, 0, 1, 1, 0）。