【Leetcode】115. Distinct Subsequences（子串的个数）（动态规划）（重点参考）

向右看齐 2022-02-16 18:21 183阅读 0赞

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

**Example 2:**

Input: S = "babgbag", T = "bag"
    Output: 5

题目大意：

找出S中能够组成T的个数。

解题思路：

动态规划，每个位置需要考虑其前一个位置的组合方式的多少。如果S中的字符与T中的字符相同，则需要其加上原先的组合，否则只需要等于上一个T中字符所产生的组合数。

注：最近好像更新了数据，表示中间的变量可能超出int。

class Solution {
    public:
        int numDistinct(string s, string t) {
            int len1 = s.length();
            int len2 = t.length();
            if(len1 == 0 || len2 == 0){
                return 0;
            }
            vector<vector<long long int>> dp(len1 + 1, vector<long long int>(len2 + 1, 0));
            dp[0][0] = 1;
            for(int i = 1; i <= len1; ++i){
                dp[i][0] = 1;
                for(int j = 1; j <= i && j <= len2; ++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[len1][len2];
        }
    };