• 115. Distinct Subsequences


    Given a string S and a string 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 1:

    Input: S = "rabbbit", T = "rabbit"
    Output: 3
    Explanation:
    
    As shown below, there are 3 ways you can generate "rabbit" from S.
    (The caret symbol ^ means the chosen letters)
    
    rabbbit
    ^^^^ ^^
    rabbbit
    ^^ ^^^^
    rabbbit
    ^^^ ^^^
    

    Example 2:

    Input: S = "babgbag", T = "bag"
    Output: 5
    Explanation:
    
    As shown below, there are 5 ways you can generate "bag" from S.
    (The caret symbol ^ means the chosen letters)
    
    babgbag
    ^^ ^
    babgbag
    ^^    ^
    babgbag
    ^    ^^
    babgbag
      ^  ^^
    babgbag
        ^^^
    class Solution {
        public int numDistinct(String s, String t) {
            int m = t.length();
            int n = s.length();
            int[][] dp = new int[m + 1][n + 1];
            
            for(int j = 0; j <= n; j++){
                    dp[0][j] = 1;
            }
            for(int i = 1; i <= m; i++){
                for(int j = 1; j <= n; j++){
                    dp[i][j] = dp[i][j - 1] + (t.charAt(i-1) == s.charAt(j-1) ? dp[i - 1][j - 1] : 0);
                }
            }
            return dp[m][n];
        }
    }

    分析如下:

    动态规划题目。

    以S ="rabbbit",T = "rabbit"为例):

      

     

    dp[i][j]表示T的从0开始长度为i的子串和S的从0开始长度为j的子串的匹配的个数。

    比如, dp[2][3]表示T中的ra和S中的rab的匹配情况。

    (1)显然,至少有dp[i][j] = dp[i][j - 1]. 

    比如, 因为T 中的"ra" 匹配S中的 "ra", 所以dp[2][2] = 1 。 显然T 中的"ra" 也匹配S中的 "rab",所以s[2][3] 至少可以等于dp[2][2]。

    (2) 如果T[i-1] == S[j-1], 那么dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0);

    比如, T中的"rab"和S中的"rab"显然匹配,

    根据(1), T中的"rab"显然匹配S中的“rabb”,所以dp[3][4] = dp[3][3] = 1, 

    根据(2),   T中的"rab"中的b等于S中的"rab1b2"中的b2, 所以要把T中的"rab"和S中的"rab1"的匹配个数累加到当前的dp[3][4]中。 所以dp[3][4] += dp[2][3] = 2;

    (3) 初始情况是
    dp[0][0] = 1; // T和S都是空串.
    dp[0][1 ... S.length() ] = 1; // T是空串,S只有一种子序列匹配。
    dp[1 ... T.length() ][0] = 0; // S是空串,T不是空串,S没有子序列匹配。
    ————————————————
    版权声明:本文为CSDN博主「feliciafay」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
    原文链接:https://blog.csdn.net/feliciafay/article/details/42959119

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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/11682193.html
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