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.
动态规划
思路
0 r a b b b i t0 1 1 1 1 1 1 1 1r 0 1 1 1 1 1 1 1a 0 0 1 1 1 1 1 1b 0 0 0 1 2 3 3 3b 0 0 0 0 1 3 3 3i 0 0 0 0 0 0 3 3t 0 0 0 0 0 0 0 3 state: dp[i][j] 表示S串中从开始位置到第i位置与T串从开始位置到底j位置匹配的子序列的个数function: dp[i][j] = dp[i][j-1] 就是说假设S已经匹配了j-1个字符,无论S[j]和T[i]是否匹配, 至少是dp[i][j-1]如果匹配(S[j]和T[i]匹配)让S[j - 1]和T[i - 1]去匹配 (由图得关系)dp[i][j] += dp[i-1][j-1]initial: f[i][0] = 0 f[0][j] = 1 空集也是subsequencesreturn dp[m][n]复杂度
时间O(mn) 空间O(mn)
代码
public int numDistinct(String s, String t) { int[][] dp = new int[t.length() + 1][s.length() + 1]; dp[0][0] = 1; for (int j = 0; j <= t.length(); j++) { dp[j][0] = 0; } for (int i = 0; i <= s.length(); i++) { dp[0][i] = 1; } for (int i = 1; i <= t.length(); i++) { for (int j = 1; j<= s.length(); j++) {dp[i][j] = dp[i][j - 1];if (s.charAt(j - 1) == t.charAt(i - 1)) { dp[i][j] += dp[i - 1][j - 1];} } } return dp[t.length()][s.length()];}