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LeetCode115 Distinct Subsequences
Given a string S and a string T, count the number of distinct subsequences of T in S. (Hard)
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
.
分析:
看题目感觉就跟LCS很像,考虑用双序列动态规划解决。
1. 状态:
dp[i][j]表示从第一个字符串前i个组成的子串转换为第二个字符串前j个组成的子串共有多少种方案。
2. 递推:
s[i - 1] == t[j - 1], 则dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
s[i - 1] != t[j - 1],则dp[i][j] = dp[i - 1][j];
3. 初始化:
dp[i][0] = 1(删除到没有字符只有一种方案)
4. 返回值:
dp[sz1 - 1][sz2 - 1]
代码:
1 class Solution { 2 public: 3 int numDistinct(string s, string t) { 4 int sz1 = s.size(), sz2 = t.size(); 5 int dp[sz1 + 1][sz2 + 1]; 6 memset(dp, 0 , sizeof(dp)); 7 for (int i = 0; i < sz1; ++i) { 8 dp[i][0] = 1; 9 } 10 for (int i = 1; i <= sz1; ++i) { 11 for (int j = 1; j <= sz2; ++j) { 12 if (s[i - 1] == t[j - 1]) { 13 dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]; 14 } 15 else { 16 dp[i][j] = dp[i - 1][j]; 17 } 18 } 19 } 20 return dp[sz1][sz2]; 21 } 22 };
LeetCode115 Distinct Subsequences