leetcode 115 Distinct Subsequences ----- java

Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

这道题是求字符串T是字符串S的字串的所有可能性的数目（不存在就是0）。

刚开始算的的时候比较暴力，所以超时了。

public class Solution {
    char[] word1;
    char[] word2;
    int result = 0;
    public int numDistinct(String s, String t) {
        int len1 = s.length(),len2 = t.length();
        if( len1 < len2 )
            return 0;
        word1 = s.toCharArray();
        word2 = t.toCharArray();
        for( int i = 0;i<=len1-len2;i++){
            if( word1[i] == word2[0])
                helper(i+1,1);
        }
        return result;

    }
    public void helper(int start1,int start2){

        if( start2 == word2.length ){
            result++;
            return ;
        }
        for( int i = start1;i< word1.length ;i++){
            if( word1[i] == word2[start2] )
                helper(i+1,start2+1);
        }

    }
}

所以用DP算法。

DP，化归为二维地图的走法问题。

r  a  b  b   i   t

1  0  0  0  0  0  0

r  1

a  1

b  1

b  1

b  1

i   1

t  1

如果当前字符相同，dp[i][j]结果等于用(dp[i-1][j-1])和(dp[i-1][j])求和

如果当前字符不同，dp[i][j] = dp[i-1][j]

public class Solution {

    public int numDistinct(String s, String t) {
        int len1 = s.length(),len2 = t.length();
        if( len1 < len2 )
            return 0;
        char[] word1 = s.toCharArray();
        char[] word2 = t.toCharArray();
        int[][] dp = new int[len1+1][len2+1];

        for( int i = 0;i<=len1;i++){
            for( int j = 0;j<=i && j<=len2;j++){
                if( i == 0 && j != 0)
                    dp[i][j] = 0;
                else if( j == 0)
                    dp[i][j] = 1;
                else if( word1[i-1] == word2[j-1] )
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
                else
                    dp[i][j] = dp[i-1][j];

            }
        }

        return dp[len1][len2];
        }

}