leetcode — distinct-subsequences

import java.util.Arrays;

/**
*
* Source : https://oj.leetcode.com/problems/distinct-subsequences/ *
*
* 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.
*/
public class DistinctSubsequences {

/**
*
* 求解的个数，使用动态规划
*
* 状态：
* i,j表示T中长度为i的prefix:T[0:i-1]，S中长度为j的prefix:S[0-j-1],S[j]第j个字符，T[i]第i个字符
* DP[i][j]表示：S[0:j]中包含T[0:i]唯一子串的个数，当j<i的时候DP[i][j] = 0
*
* 递推公式：
* 当S[j] != T[i]的时候
* DP[i+1][j+1] = DP[i+1][j]，含义是当前字符不相等的时候，S[j+1]包含T[i+1]的个数就是S[j]包含T[i+1]的个数
*
* 当S[j] == T[i]的时候
* DP[i+1][j+1] = DP[i+1][j] + DP[i][j]，含义是当前字符相等的时候，S[j+1]包含T[i+1]的个数就是S[j]包含T[i+1]的个数加上S[j]包含T[i]的个数
*
* 计算方向和起始状态：
* DP[i][j]
* DP[i+1][j],DP[i+1[j+1]
* 从上到下，从左到右
*
* 第0行：1
* 第0列：0
*
*
* @param S
* @param T
* @return
*/
public int distinctSequences (String S, String T) {
int[][] dp = new int[T.length()+1][S.length()+1];
for (int i = 0; i <= T.length(); i++) {
dp[i][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 = i; j <= S.length(); j++) {
if (S.charAt(j-1) == T.charAt(i-1)) {
dp[i][j] = dp[i][j-1] + dp[i-1][j-1];
} else {
dp[i][j] = dp[i][j-1];
}
}
}
return dp[T.length()][S.length()];
}

/**
* 优化DP占用空间，因为递推的时候只需要dp[i][j-1],dp[i-1][j-1]
* 也就是当前矩阵左上角的值和左面的值，使用滚动数组优化空间
*
* @param S
* @param T
* @return
*/
public int distinctSequences1 (String S, String T) {
int[] dp = new int[S.length()+1];
Arrays.fill(dp, 1);
for (int i = 1; i <= T.length(); i++) {
int upLeft = dp[0];
dp[0] = 0;
for (int j = 1; j <= S.length(); j++) {
// 相当于记下dp[i-1][j-1]
int temp = dp[j];
// 相当于dp[i][j-1]
dp[j] = dp[j-1];
if (S.charAt(j-1) == T.charAt(i-1)) {
dp[j] += upLeft;
}
upLeft = temp;
}
}
return dp[S.length()];
}

public static void main(String[] args) {
DistinctSubsequences subsequences = new DistinctSubsequences();
System.out.println(subsequences.distinctSequences("rabbbit", "rabbit") + "------3");
System.out.println(subsequences.distinctSequences1("rabbbit", "rabbit") + "------3");
}
}