leetcode-115. Distinct Subsequences

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1、原题链接

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).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

 

链接:https://www.nowcoder.com/questionTerminal/ed2923e49d3d495f8321aa46ade9f873
来源:牛客网
2、思路

 *  思路:dp题。
 *  状态定义:dp[i][j]代表s[0~j - 1]中T[0~i - 1]不同子串的个数。
 *  递推关系式:S[j-1]!= T[i-1]:  DP[i][j] = DP[i][j - 1] (不选择S中的s[j-1]字符)
 *              S[j-1]==T[i-1]: DP[i][j] = DP[i-1][j-1](选择S中的s[j-1]字符) + DP[i][j - 1](不选择S中的s[j-1]字符)
 *  初始状态:第0列:DP[i][0] = 1,第0行:DP[0][j] = 0
 */
当s[j - 1] = t[i - 1]时,s[j - 1]要分两种情况,选或者不选本字符,所以是加和的关系。当s[j - 1] != t[i - 1]时,选择本字符的情况与不选择本字符情况一样dp[i][j - 1]
 
3、代码实现
 1 class Solution {
 2 public:
 3     int numDistinct(string s, string t) {
 4       int ss = s.size() + 1;
 5         int st = t.size() + 1;
 6         int dp[st][ss];
 7         memset(dp, 0, st* ss* sizeof(int));
 8         for (int i = 0; i < ss; i++) {
 9             dp[0][i] = 1;
10         }
11         for (int i = 1; i < st; i++) {
12             for (int j = 1; j < ss; j ++) {
13                 if (s[j - 1] == t[i - 1]) {
14                     dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1];
15                 } else {
16                     dp[i][j] = dp[i][j - 1]; 
17                 }
18             }
19         }
20         return dp[st - 1][ss - 1];
21     }
22 };

4、优化

在使用中不难发现该dp二维数组可以降维,注意改变数组元素值时由后往前

 1 class Solution {
 2 public:
 3 int numDistinct(string S, string T) {
 4     int len=T.size();
 5     vector<int> array(len+1);
 6     array[0]=1;
 7     for(int i=1;i<S.size()+1;i++){
 8         for(int j=min(i,len);j>0;j--){
 9             if(S[i-1]==T[j-1])
10                 array[j]=array[j]+array[j-1];
11         }
12     }
13     return array[len];
14     }
15 };

5、api

6、参考牛客网 https://www.nowcoder.com/questionTerminal/ed2923e49d3d495f8321aa46ade9f873

 



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