标签：character rabbit code ++ logs not one state 数组
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). Have you met this question in a real interview? Yes Example Given S = "rabbbit", T = "rabbit", return 3. Challenge Do it in O(n2) time and O(n) memory. O(n2) memory is also acceptable if you do not know how to optimize memory.
动归: 状态的加法
分完情况后根据遍历到的点的S的当前字母匹不匹配T的当前字母的问题来进行状态转移:
public int numDistinct(String S, String T) {
// write your code here
//state
int m = S.length();
int n = T.length();
// initialize
int[][] f = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
f[i][0]= 1;
}
for (int j = 1; j <= n; j++) {
f[0][j] = 0;
}
//function
for (int i = 1; i<= m; i++) {
for (int j = 1; j <= n; j++) {
if (S.charAt(i - 1) == T.charAt(j - 1)) {
f[i][j] = f[i- 1][j - 1] + f[i - 1][j];
} else {
f[i][j] = f[i - 1][j];
}
}
}
return f[m][n];
}
一维数组. 逆序以保证 f[i][j] = f[i- 1][j - 1] + f[i - 1][j]; 加的是上一层的i- 1
public class Solution {
public int numDistinct(String S, String T) {
if (S==null || T==null || S.length() < T.length()) return 0;
int[] res = new int[T.length()+1];
res[0] = 1;
for (int i=1; i<=S.length(); i++) {
for (int j=T.length(); j>0; j--) {
res[j] = S.charAt(i-1) == T.charAt(j-1)? res[j-1] + res[j] : res[j];
}
}
return res[T.length()];
}
}
标签：character rabbit code ++ logs not one state 数组
原文地址：http://www.cnblogs.com/apanda009/p/7291410.html
