**[Lintcode]Distinct Subsequences

最新推荐文章于 2019-07-09 16:33:37 发布

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最新推荐文章于 2019-07-09 16:33:37 发布

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分类专栏：算法文章标签： lintcode

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本文介绍了一种使用二维动态规划解决子序列计数问题的方法，该方法可在O(n²)的时间复杂度和O(n)的空间复杂度下运行，提供了一个Java实现示例。

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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).

Example

Given S = "rabbbit", T = "rabbit", return 3.

Challenge

Do it in O(n2) time and O(n) memory.

O(n2) memory is also acceptable if you do not know how to optimize memory.

使用二维DP。res[i][j]代表截止到T中第i个字符和S中第j个字符，可能匹配的字符串的数量。其中公式为：

if s[i] = t[j], res[i][j] = res[i][j - 1] + res[i - 1][j - 1]

else res[i][j] = res[i][j - 1]

当i=j时，res[i][j]等于，第i个字符对应到第j个字符后，匹配字符串的数量加上第i个字符对应到1～j-1的某个字符后，匹配字符串的数量。

例子中的匹配结果如图：

    0    r    a    b    b    b    i    t
0    1    1    1    1    1    1    1    1
r    0    1    1    1    1    1    1    1
a    0    0    1    1    1    1    1    1
b    0    0    0    1    2    3    3    3
b    0    0    0    0    1    3    3    3
i    0    0    0    0    0    0    3    3
t    0    0    0    0    0    0    0    3

public class Solution {
    /**
     * @param S, T: Two string.
     * @return: Count the number of distinct subsequences
     */
    public int numDistinct(String S, String T) {
        int[][] res = new int[T.length() + 1][S.length() + 1];
        
        //if s[i] = t[j], res[i][j] = res[i][j - 1] + res[i - 1][j - 1]
        //else res[i][j] = res[i][j - 1]
        
        int lengthS = S.length(), lengthT = T.length();
        //"" equals to all string
        for(int i = 0; i <= lengthS; i++) res[0][i] = 1;
        
        for(int i = 1; i <= lengthT; i++) {
            for(int j = i; j <= lengthS; j++) {
                if(S.charAt(j - 1) == T.charAt(i - 1)) {
                    res[i][j] = res[i][j - 1] + res[i - 1][j - 1];
                } else {
                    res[i][j] = res[i][j - 1];
                }
            }
        }
        return res[lengthT][lengthS];
    }
}