lintcode: 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).

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

设矩阵transArray，其中元素transArray[i][j]为S[0,…,i] （这里指的是前0到i个字母）到T[0,…,j]有多少种转换方式。

问题就转为从左上角只能走对角（匹配）或者往下（删除字符），到右下角一共有多少种走法。

transArray[i][0]初始化为1的含义是：任何长度的S，如果转换为空串，那就只有删除全部字符这1种方式。

当S[i-1]==T[j-1]，说明可以从transArray[i-1][j-1]走对角到达transArray[i][j]（S[i-1]匹配T[j-1]），此外还可以从transArray[i-1][j]往下到达transArray[i][j]（删除S[i-1]）

参考http://www.cnblogs.com/ganganloveu/p/3836519.html

class Solution {public:        /**     * @param S, T: Two string.     * @return: Count the number of distinct subsequences     */    int numDistinct(string &S, string &T) {        // write your code here        int lenS=S.size();        int lenT=T.size();        vector<vector<int> >num(lenS+1,vector<int>(lenT+1,0));        for(int i=0;i<=lenS;i++){            num[i][0]=1;        }        for(int i=1;i<=lenS;i++){            for(int j=1;j<=lenT;j++){                if(S[i-1]==T[j-1]){                    num[i][j]=num[i-1][j-1]+num[i-1][j];                }else{                    num[i][j]=num[i-1][j];                }            }        }        return num[lenS][lenT];    }};

一维数组即可

class Solution {public:        /**     * @param S, T: Two string.     * @return: Count the number of distinct subsequences     */    int numDistinct(string &S, string &T) {        // write your code here        int lenS=S.size();        int lenT=T.size();        //vector<vector<int> >num(lenS+1,vector<int>(lenT+1,0));        vector<int> num(lenT+1,0);        num[0]=1;        /*        for(int i=0;i<=lenS;i++){            num[i][0]=1;        }        */        for(int i=1;i<=lenS;i++){            /*注意这里保存上一层的值*/            vector<int> oldNum(num);            for(int j=1;j<=lenT;j++){                if(S[i-1]==T[j-1]){                    num[j]=oldNum[j-1]+oldNum[j];                }            }        }        return num[lenT];    }};