标签：pre 求和 and rabbit bsp array let ati origin
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.
这道题是求字符串T是字符串S的字串的所有可能性的数目（不存在就是0）。
刚开始算的的时候比较暴力，所以超时了。
public class Solution { char[] word1; char[] word2; int result = 0; public int numDistinct(String s, String t) { int len1 = s.length(),len2 = t.length(); if( len1 < len2 ) return 0; word1 = s.toCharArray(); word2 = t.toCharArray(); for( int i = 0;i<=len1-len2;i++){ if( word1[i] == word2[0]) helper(i+1,1); } return result; } public void helper(int start1,int start2){ if( start2 == word2.length ){ result++; return ; } for( int i = start1;i< word1.length ;i++){ if( word1[i] == word2[start2] ) helper(i+1,start2+1); } } }
所以用DP算法。
DP，化归为二维地图的走法问题。
r a b b i t
1 0 0 0 0 0 0
r 1
a 1
b 1
b 1
b 1
i 1
t 1
如果当前字符相同，dp[i][j]结果等于用(dp[i-1][j-1])和(dp[i-1][j])求和
如果当前字符不同，dp[i][j] = dp[i-1][j]
public class Solution { public int numDistinct(String s, String t) { int len1 = s.length(),len2 = t.length(); if( len1 < len2 ) return 0; char[] word1 = s.toCharArray(); char[] word2 = t.toCharArray(); int[][] dp = new int[len1+1][len2+1]; for( int i = 0;i<=len1;i++){ for( int j = 0;j<=i && j<=len2;j++){ if( i == 0 && j != 0) dp[i][j] = 0; else if( j == 0) dp[i][j] = 1; else if( word1[i-1] == word2[j-1] ) dp[i][j] = dp[i-1][j-1] + dp[i-1][j]; else dp[i][j] = dp[i-1][j]; } } return dp[len1][len2]; } }
leetcode 115 Distinct Subsequences ----- java
标签：pre 求和 and rabbit bsp array let ati origin
原文地址：http://www.cnblogs.com/xiaoba1203/p/6021400.html
