[LeetCode] 115. Distinct Subsequences
2017-03-03 13:45
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[LeetCode] 115. Distinct Subsequences
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.
找到T在S中有几种子序列表示,如:
S = “rabbbit”, T = “rabbit”
[
rabb b it
rab b bit
ra b bbit
]
三种
动规解决,找到转移方程:
if (s[i] == t[j]) {
dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
} else {
dp[i][j] = dp[i-1][j];
}
dp[i-1][j-1] 表示s[0:i-1]已经有这么多个t[0:j-1]的子序列。
dp[i-1][j] 表示s[0:i-1]已经有这么多个t[0:j]的子序列。
如果s[i] == t[j], 可以在dp[i-1][j]数量的基础上新增dp[i-1][j-1]个子序列。
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<int>> dp(len1, vector<int>(len2, 0)); if (s[0] == t[0]) dp[0][0] = 1; for (int i=1; i<len1; ++i) { if (s[i] == t[0]) dp[i][0] = dp[i-1][0] + 1; else dp[i][0] = dp[i-1][0]; } for (int i=1; i<len1; ++i) { for (int j=1; j<len2; ++j) { if (s[i] == t[j]) dp[i][j] = dp[i-1][j-1] + dp[i-1][j]; else dp[i][j] = dp[i-1][j]; } } return dp[len1-1][len2-1]; } };
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