LeetCode OJ

这道题采用动态规划，可是我一开始没有想到。后来参考了discuss中前辈的代码和思路，才想通的。 方法二是因为每一步只和上一步的内容相关，所以可以只用O（n）的空间复杂度。

下面是AC代码：

/**
     * Solution DP
     * we keep a m*n matrix and scanning through string T,
     * p[i][j] means the number of distinct subsequence of S(0...j) equal to T(0...i) 
     * p[i][j] =  p[i][j-1]   discard s[j]
     *         +  0           if s[j] != t[i]
     *         +  p[i-1][j-1] if s[j] == t[i]
     * @param S
     * @param T
     * @return
     */
    public int numDistinct1(String S, String T){
        if(S.length()<T.length())
            return 0;
        int[][] p = new int[T.length()][S.length()];
        //for convenient
        char[] Sc = S.toCharArray();
        char[] Tc = T.toCharArray();
        
        p[0][0] = Tc[0] == Sc[0] ? 1:0;
        
        for(int i=1;i<S.length();i++)
            p[0][i] = p[0][i-1]+(Tc[0] == Sc[i]?1:0);
        
        for(int j = 1;j<T.length();j++)
            p[j][0] = 0;
        
        for(int i=1;i<T.length();i++){
            for(int j=1;j<S.length();j++){
                p[i][j] = p[i][j-1]+(Tc[i] == Sc[j] ? p[i-1][j-1]:0);
            }
        }
        return p[T.length()-1][S.length()-1];
    }
    /**
     * O(n) space complexity solution
     * @param S
     * @param T
     * @return
     */
    public int numDistinct2(String S,String T){
        if(S.length()<T.length())
            return 0;
        int[] p = new int[S.length()];
        //for convenient
        char[] Sc = S.toCharArray();
        char[] Tc = T.toCharArray();
       
        p[0] = Tc[0] == Sc[0] ? 1:0;
        
        for(int i=1;i<S.length();i++)
            p[i] = p[i-1]+(Tc[0] == Sc[i]?1:0);
        
        for(int i=1;i<T.length();i++){
            for(int j=S.length()-1;j>=0;j--){
                p[j] = p[j] + (Tc[i] == Sc[j]? p[j-1] :0);
            }
        }
        return p[S.length()-1];
       
    }