LeetCode: Distinct Subsequences 解题报告

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.

SOLUTION 1（AC）:

现在这种DP题目基本都是5分钟AC咯。主页君引一下别人的解释咯：

http://blog.csdn.net/fightforyourdream/article/details/17346385?reload#comments

http://blog.csdn.net/abcbc/article/details/8978146

引自以上的解释：

遇到这种两个串的问题，很容易想到DP。但是这道题的递推关系不明显。可以先尝试做一个二维的表int[][] dp，用来记录匹配子序列的个数（以S ="rabbbit",T = "rabbit"为例）：

r a b b b i t

1 1 1 1 1 1 1 1

r 0 1 1 1 1 1 1 1

a 0 0 1 1 1 1 1 1

b 0 0 0 1 2 3 3 3

b 0 0 0 0 1 3 3 3

i 0 0 0 0 0 0 3 3

t 0 0 0 0 0 0 0 3

从这个表可以看出，无论T的字符与S的字符是否匹配，dp[i][j] = dp[i][j - 1].就是说，假设S已经匹配了j - 1个字符，得到匹配个数为dp[i][j - 1].现在无论S[j]是不是和T[i]匹配，匹配的个数至少是dp[i][j - 1]。除此之外，当S[j]和T[i]相等时，我们可以让S[j]和T[i]匹配，然后让S[j - 1]和T[i - 1]去匹配。所以递推关系为：

dp[0][0] = 1; // T和S都是空串.

dp[0][1 ... S.length() - 1] = 1; // T是空串，S只有一种子序列匹配。

dp[1 ... T.length() - 1][0] = 0; // S是空串，T不是空串，S没有子序列匹配。

dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0).1 <= i <= T.length(), 1 <= j <= S.length()

这道题可以作为两个字符串DP的典型：

两个字符串：

先创建二维数组存放答案，如解法数量。注意二维数组的长度要比原来字符串长度+1，因为要考虑第一个位置是空字符串。

然后考虑dp[i][j]和dp[i-1][j],dp[i][j-1],dp[i-1][j-1]的关系，如何通过判断S.charAt(i)和T.charAt(j)的是否相等来看看如果移除了最后两个字符，能不能把问题转化到子问题。

最后问题的答案就是dp[S.length()][T.length()]

还有就是要注意通过填表来找规律。

注意：循环的时候，一定要注意i的取值要到len，这个出好几次错了。

 public class Solution {
     public int numDistinct(String S, String T) {
         if (S == null || T == null) {
             return 0;
         }

         int lenS = S.length();
         int lenT = T.length();

         if (lenS < lenT) {
             return 0;
         }

         int[][] D = new int[lenS + 1][lenT + 1];

         // BUG 1: forget to use <= instead of <....
         for (int i = 0; i <= lenS; i++) {
             for (int j = 0; j <= lenT; j++) {
                 // both are empty.
                 if (i == 0 && j == 0) {
                     D[i][j] = 1;
                 } else if (i == 0) {
                     // S is empty, can't form a non-empty string.
                     D[i][j] = 0;
                 } else if (j == 0) {
                     // T is empty. S is not empty.
                     D[i][j] = 1;
                 } else {
                     D[i][j] = 0;
                     // keep the last character of S.
                     if (S.charAt(i - 1) == T.charAt(j - 1)) {
                         D[i][j] += D[i - 1][j - 1];
                     }

                     // discard the last character of S.
                     D[i][j] += D[i - 1][j];
                 }
             }
         }

         return D[lenS][lenT];
     }
 }

运行时间：

Submit Time	Status	Run Time	Language
13 minutes ago	Accepted	432 ms	java

SOLUTION 2:

递归解法也写一下，蛮简单的：

但是这个解法过不了，TLE了。

 // SOLUTION 2: recursion version.
     public int numDistinct(String S, String T) {
         if (S == null || T == null) {
             return 0;
         }

         return rec(S, T, 0, 0);
     }

     public int rec(String S, String T, int indexS, int indexT) {
         int lenS = S.length();
         int lenT = T.length();

         // base case:
         if (indexT >= lenT) {
             // T is empty.
             return 1;
         }

         if (indexS >= lenS) {
             // S is empty but T is not empty.
             return 0;
         }

         int sum = 0;
         // use the first character in S.
         if (S.charAt(indexS) == T.charAt(indexT)) {
             sum += rec(S, T, indexS + 1, indexT + 1);
         }

         // Don't use the first character in S.
         sum += rec(S, T, indexS + 1, indexT);

         return sum;
     }

SOLUTION 3:

递归加上memory记忆之后，StackOverflowError. 可能还是不够优化。确实递归层次太多。

Runtime Error Message:	Line 125: java.lang.StackOverflowError
Last executed input:	"zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

 // SOLUTION 3: recursion version with memory.
     public int numDistinct(String S, String T) {
         if (S == null || T == null) {
             return 0;
         }

         int lenS = S.length();
         int lenT = T.length();

         int[][] memory = new int[lenS + 1][lenT + 1];
         for (int i = 0; i <= lenS; i++) {
             for (int j = 0; j <= lenT; j++) {
                 memory[i][j] = -1;
             }
         }

         return rec(S, T, 0, 0, memory);
     }

     public int rec(String S, String T, int indexS, int indexT, int[][] memory) {
         int lenS = S.length();
         int lenT = T.length();

         // base case:
         if (indexT >= lenT) {
             // T is empty.
             return 1;
         }

         if (indexS >= lenS) {
             // S is empty but T is not empty.
             return 0;
         }

         if (memory[indexS][indexT] != -1) {
             return memory[indexS][indexT];
         }

         int sum = 0;
         // use the first character in S.
         if (S.charAt(indexS) == T.charAt(indexT)) {
             sum += rec(S, T, indexS + 1, indexT + 1);
         }

         // Don't use the first character in S.
         sum += rec(S, T, indexS + 1, indexT);

         // record the solution.
         memory[indexS][indexT] = sum;
         return sum;
     }

SOLUTION 4 （AC）:

参考了http://blog.csdn.net/fightforyourdream/article/details/17346385?reload#comments的代码后，发现递归过程找解的过程可以优化。我们不需要沿用DP的思路

而应该与permutation之类差不多，把当前可能可以取的解都去尝试一次。就是在S中找到T的首字母，再进一步递归。

Submit Time	Status	Run Time	Language
0 minutes ago	Accepted	500 ms	java

 // SOLUTION 4: improved recursion version
     public int numDistinct(String S, String T) {
         if (S == null || T == null) {
             return 0;
         }

         int lenS = S.length();
         int lenT = T.length();

         int[][] memory = new int[lenS + 1][lenT + 1];
         for (int i = 0; i <= lenS; i++) {
             for (int j = 0; j <= lenT; j++) {
                 memory[i][j] = -1;
             }
         }

         return rec4(S, T, 0, 0, memory);
     }

     public int rec4(String S, String T, int indexS, int indexT, int[][] memory) {
         int lenS = S.length();
         int lenT = T.length();

         // base case:
         if (indexT >= lenT) {
             // T is empty.
             return 1;
         }

         if (indexS >= lenS) {
             // S is empty but T is not empty.
             return 0;
         }

         if (memory[indexS][indexT] != -1) {
             return memory[indexS][indexT];
         }

         int sum = 0;
         for (int i = indexS; i < lenS; i++) {
             // choose which character in S to choose as the first character of T.
             if (S.charAt(i) == T.charAt(indexT)) {
                 sum += rec4(S, T, i + 1, indexT + 1, memory);
             }
         }

         // record the solution.
         memory[indexS][indexT] = sum;
         return sum;
     }

SOLUTION 5:

在SOLUTION 4的基础之上，把记忆体去掉之后，仍然是TLE

Time Limit ExceededMore Details

Last executed input:

"daacaedaceacabbaabdccdaaeaebacddadcaeaacadbceaecddecdeedcebcdacdaebccdeebcbdeaccabcecbeeaadbccbaeccbbdaeadecabbbedceaddcdeabbcdaeadcddedddcececbeeabcbecaeadddeddccbdbcdcbceabcacddbbcedebbcaccac", "ceadbaa"

 // SOLUTION 5: improved recursion version without memory.
     public int numDistinct(String S, String T) {
         if (S == null || T == null) {
             return 0;
         }

         return rec5(S, T, 0, 0);
     }

     public int rec5(String S, String T, int indexS, int indexT) {
         int lenS = S.length();
         int lenT = T.length();

         // base case:
         if (indexT >= lenT) {
             // T is empty.
             return 1;
         }

         if (indexS >= lenS) {
             // S is empty but T is not empty.
             return 0;
         }

         int sum = 0;
         for (int i = indexS; i < lenS; i++) {
             // choose which character in S to choose as the first character of T.
             if (S.charAt(i) == T.charAt(indexT)) {
                 sum += rec5(S, T, i + 1, indexT + 1);
             }
         }

         return sum;
     }

总结:

大家可以在SOLUTION 1和SOLUTION 4两个选择里用一个就好啦。

http://blog.csdn.net/fightforyourdream/article/details/17346385?reload#comments

这道题可以作为两个字符串DP的典型：

两个字符串：

先创建二维数组存放答案，如解法数量。注意二维数组的长度要比原来字符串长度+1，因为要考虑第一个位置是空字符串。

然后考虑dp[i][j]和dp[i-1][j],dp[i][j-1],dp[i-1][j-1]的关系，如何通过判断S.charAt(i)和T.charAt(j)的是否相等来看看如果移除了最后两个字符，能不能把问题转化到子问题。

最后问题的答案就是dp[S.length()][T.length()]

还有就是要注意通过填表来找规律。

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/dp/NumDistinct.java