LintCode 118: Distinct Subsequences (DP经典好题)

原创于 2019-07-09 16:33:37 发布 · 289 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#LintCode

algorithm-design 专栏收录该内容

817 篇文章

订阅专栏

博客围绕118. Distinct Subsequences问题展开，介绍了计算字符串S中等于字符串T的不同子序列数量的问题，给出示例和挑战要求，需在O(n^2)时间和O(n)内存完成，还给出了使用动态规划（DP）的解法思路及代码。

118. Distinct Subsequences

中文English

Given two strings S and T. Count the number of distinct subsequences of S which equals T.

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

Example 1:

Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation: You could remove any 'b' in S, so there are 3 ways to get T.

Example 2:

Input: S = "abcd", T = ""
Output: 1
Explanation: There is only 1 way to get T - remove all chars in S.

Challenge

Do it in O(n^2 time and O(n) memory.

O(n^2n) memory is also acceptable if you do not know how to optimize memory.

Input test data (one parameter per line)How to understand a testcase?

解法1：DP

关键思路：S[i] vs T[j]。因为不管怎么样，我们可以丢弃掉S[i]，这样问题可以简化到S[i-1] vs T[j]。

而当S[i-1]==T[j-1]时，我们还可以将问题简化到S[i-1] vs T[j-1]。

代码如下:

class Solution {
public:
    /**
     * @param S: A string
     * @param T: A string
     * @return: Count the number of distinct subsequences
     */
    int numDistinct(string &S, string &T) {
        int M = S.size();
        int N = T.size();
        if (M == 0 || M < N) return 0;
        
        vector<vector<int>> dp(M + 1, vector<int>(N + 1, 0));
        
        for (int i = 0; i <= M; ++i) {
            dp[i][0] = 1; //delete any char can get empty
        }

        for (int i = 1; i <= M; ++i) {
            for (int j = 1; j <= N; ++j) {
                dp[i][j] = dp[i - 1][j];
                if (S[i - 1] == T[j - 1]) {
                    dp[i][j] += dp[i - 1][j - 1];
                }
                
            }
        }

        return dp[M][N];
    }
};