LCS空间优化&01背包利用滚动数组优化空间&逆序问题

 没法用逆序做，只能用顺序做。因为这里涉及到3个状态量，即左，上，左上三个状态量，用temp变量保存左上角的变量。空间复杂度O(n+1）+1。

但是01背包问题却可以用逆序做，因为只涉及两个状态量：上面的和左上角的。 

 

//LCS只用一个数组+一个prev变量的解法。其中backup变量可以省去 ，用curr[0]代替。数组长度取X和Y的长度的最小值。

#include <iostream>
#include <string>
using namespace std;

// Space optimized function to find length of Longest Common Subsequence
// of substring X[0..m-1] and Y[0..n-1]
int LCSLength(string X, string Y)
{
	int m = X.length(), n = Y.length();

	// allocate storage for one-dimensional array curr
	int curr[n + 1], prev;

	// fill the lookup table in bottom-up manner
	for (int i = 0; i <= m; i++)
	{
		prev = curr[0];
		for (int j = 0; j <= n; j++)
		{
			int backup = curr[j];

			if (i == 0 || j == 0)
				curr[j] = 0;
			else
			{
				// if current character of X and Y matches
				if (X[i - 1] == Y[j - 1])
					curr[j] = prev + 1;

				// else if current character of X and Y don't match
				else
					curr[j] = max(curr[j], curr[j - 1]);
			}
			prev = backup;
		}
	}

	// LCS will be last entry in the lookup table
	return curr[n];
}

// Longest Common Subsequence problem space optimized solution
int main()
{
	string X = "XMJYAUZ", Y = "MZJAWXU";

	// pass smaller string as second argument to LCSLength()
	if (X.length() > Y.length())
		cout << "The length of LCS is " << LCSLength(X, Y);
	else
		cout << "The length of LCS is " << LCSLength(Y, X);

	return 0;
}

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LCS的递归写法：