Given a string S and a string T, count the number of distinct subsequences of S which equals T.
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.
一刷
用dp来求解。构造二维数组dp[s.length()+1][t.length()+1], dp[i][0]=1;(当t为[]时),如果s.charAt(i-1) == t.charAt(j-1), dp[i][j] = dp[i-1][j-1] + dp[i-1][j], 如果不等, dp[i][j] = dp[i-1][j]
public class Solution {
public int numDistinct(String s, String t) {
if(s == null || t == null) return 0;
int[][] dp = new int[s.length() + 1][t.length() + 1];
for(int i=0; i<=s.length(); i++){
dp[i][0] = 1;
}
for(int i=1; i<=s.length(); i++){
for(int j=1; j<=t.length(); j++){
dp[i][j] = dp[i-1][j];
if(s.charAt(i-1) == t.charAt(j-1))
dp[i][j] += dp[i-1][j-1];
}
}
return dp[s.length()][t.length()];
}
}
二刷
空间复杂度O(m)
由于我们仅用到了dp[i-1][j-1], dp[i-1][j],考虑用一维数组替换二维数组
对于i, j, 未修改dp[j]时,此时dp[j]表示dp[i-1][j], 对于下一个j, 表示为dp[i-1][j-1]
public class Solution {
public int numDistinct(String s, String t) {
int n = s.length(), m = t.length();
int[] cur = new int[m+1];
cur[0] = 1;
for(int i=1; i<=n; i++){
int prev = 1;
for(int j=1; j<=m; j++){
int temp = cur[j];
if(s.charAt(i-1) == t.charAt(j-1)) cur[j] += prev;
prev = temp;
}
}
return cur[m];
}
}
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