[数据结构与算法]Leetcode——115. Distinct Subsequences(C++)

Leetcode——115. Distinct Subsequences(C++)
Given two strings s and t, return the number of distinct subsequences of s which equals t.
A string’s subsequence is a new string formed from the original string by deleting some (can be none) of the characters without disturbing the remaining characters’ relative positions. (i.e., “ACE” is a subsequence of “ABCDE” while “AEC” is not).
The test cases are generated so that the answer fits on a 32-bit signed integer.

Example 1:
Input: s = “rabbbit”, t = “rabbit”
Output: 3
Explanation:
As shown below, there are 3 ways you can generate “rabbit” from S.
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.
babgbag
babgbag
babgbag
babgbag
babgbag

Constraints:
1 <= s.length, t.length <= 1000
s and t consist of English letters.

解题思路：
动态规划。
Example 2:
Input: s = “babgbag”, t = “bag”

代码：

class Solution {
public:
    int numDistinct(string s, string t) {
        int ssiz=s.size();
        int tsiz=t.size();
        vector<vector<int>>dp(tsiz+1,vector<int>(ssiz+1));
        dp[0][0]=0;
        for(int i=0;i<=tsiz;i++)dp[i][0]=0;
        for(int j=0;j<=ssiz;j++)dp[0][j]=1;
        int mood=1000000007;  //为什么我将第一行置零（我知道第一行置零答案不对，但是我第二行另算了啊），然后第二行另算就是出不来解，明明稿纸演算不错！！！
        for(int i=1;i<=tsiz;i++)
        {
            for(int j=1;j<=ssiz;j++)
            {
                if(s[j-1]==t[i-1])
                {
                    dp[i][j]=(dp[i-1][j-1]+dp[i][j-1])%mood;
                }
                else
                    dp[i][j]=dp[i][j-1];
            }
        
        }
        return dp[tsiz][ssiz];
    }
};

后记：
思路是这么个思路，但是为什么我一开始以第二行（也就是是t[0]行）单独计算时，结果就是不对，手算时倒没问题；
还有就是整数溢出问题。。。
本人好菜。。。