[LeetCode]Distinct Subsequences

最新推荐文章于 2017-07-04 10:43:23 发布

最新推荐文章于 2017-07-04 10:43:23 发布 · 100 阅读

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本文介绍了一种使用动态规划方法来解决字符串匹配问题的具体实现。该方法能够计算一个字符串作为另一个字符串子序列出现的不同方式的数量。

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.

方法：dp

a[i][j]表示T中前j个在S的前i个出现的个数。

则

if S[i] != T[j]

a[i][j] = a[i-1][j]，相当于如果S中前i-1个可以表示T中前j个的话，那么多一个S[i]也不会影响。

if S[i] == T[j]

a[i][j] = a[i-1][j] + a[i-1][j-1]

class Solution {
public:
    int numDistinct(string S, string T) {
        int* a = new int[T.size()];
        memset(a, 0, sizeof(int) * T.size());
        
        for (int i = 0; i < S.size(); i++) {
            for (int j = T.size()-1; j >= 0; j--) {
                if (S[i] == T[j]) { 
                    if (j==0) a[j] += 1;
                    else a[j] += a[j-1];
                }
            }
        }
        return a[T.size()-1];
    }
};