【一天一道LeetCode】#115. Distinct Subsequences
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一天一道LeetCode
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(一)题目
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.
(二)解题
题目大意:给定字符串T和S,求S的子串中有多少与T相等。
解题思路:S的子串代表在S中删除一些字符组成的字符串,那么可以很容易想到用递归来解决。
如果当前s[i]==s[j],有两种情况,选择该字母(i++,j++)和跳过该字母(i++)
如果s[i]!=s[j],则直接i++.
具体见代码:
class Solution {public: int numDistinct(string s, string t) { if(s.length()==0||t.length()==0) return 0; int num =0; dpDistinct(s,t,0,0,num); return num; } void dpDistinct(string& s, string& t , int i , int j,int& num) { if(j==t.length()){num++;return;} if(s[i]==t[j]) { //choose this words if(i<s.length()&&j<t.length()) dpDistinct(s, t , i+1, j+1,num); //not choose this words if(i<s.length()) dpDistinct(s, t , i+1, j,num); } else //Don't equal if(i<s.length()) dpDistinct(s, t , i+1, j,num); }};
然后……直接超时了。
观察上述代码,发现有很多重复的判断,因此,可以采用动态规划的思想。
开始找状态转移方程。dp[i][j]用来表示s中j之前的子串subs中有多少个不同的subt,其中subt为t中i之前的字符组成的子串。
首先,初始化dp,令dp[0][j]都等于1,因为s中删除所有的字符都能组成空串。
如果t[i] == s[j],那么dp[i][j] = dp[i][j-1]+dp[i-1][j-1]
否则,dp[i][j] = dp[i][j-1]
举例说明一下:s为abbbc,t为abc
具体解释看代码:
class Solution {public: int numDistinct(string s, string t) { vector<vector<int>> dp; for(int i = 0 ; i < t.length()+1;i++) { vector<int> temp(s.length()+1,0); dp.push_back(temp); } for(int i = 0 ; i < s.length()+1;i++) dp[0][i]=1;//初始化dp for(int i = 1 ; i < t.length()+1; i++) { for(int j = 1 ; j < s.length()+1 ; j++) { if(s[j-1]==t[i-1]) { dp[i][j] = dp[i-1][j-1]+dp[i][j-1];//状态转移方程 } else dp[i][j] = dp[i][j-1]; } } return dp[t.length()][s.length()];//返回 }};
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