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
.
https://oj.leetcode.com/problems/distinct-subsequences/
思路:DP。dp[i][j]代表从s[0..i]中删除几个字符得到t[0..j]的不同删除方法数。
- 初始化:如果t为空,任意的s删除到t都是1种。
- 如果S[i] = T[j], dp[i][j] = dp[i-1][j-1]+dp[i-1][j]
- 如果S[i] 不等于 T[j], dp[i][j] = dp[i-1][j]
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
public class Solution { public int numDistinct(String S, String T) { int m = S.length(); int n = T.length(); int[][] dp = new int[m + 1][n + 1]; for (int i = 0; i <= m; i++) dp[i][0] = 1; for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { if (S.charAt(i - 1) == T.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]; else dp[i][j] = dp[i - 1][j]; } } return dp[m][n]; } public static void main(String[] args) { String S = "rabbbit"; String T = "rabbit"; System.out.println(new Solution().numDistinct(S, T)); } }
第二遍记录:画表格辅助分析。
递推的时候判断s[i] 和 t[j]的关系,
如果不相等, dp[i][j]=dp[i-1][j],考虑 rab -> ra的情况,即rab->ra的情况数与 ra->ra是一样的(只不过牵着每种情况删除时多=删一个b)。
如果相等,还需加上dp[i-1][j-1]的情况,考虑rabb-> rab的情况,除了上述情况之外,还需要多考虑一种 rab->ra的情况。
第三遍记录:
参考:
http://www.programcreek.com/2013/01/leetcode-distinct-subsequences-total-java