leetcode，leetcodeoj

leetcode，leetcodeoj

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.

//這裡可以用dfs,但是，TLE.所以，採用dp來解決.//假設dp[i][j]的狀態表示為字串i變換到字串j的方法的數量，那麼，動態轉移方程為：// 1) S[i-1] == T[j-1] --> dp[i][j] = dp[i-1][j-1] + dp[i-1][j]// 2) S[i-1] != T[j-1] --> dp[i][j] = dp[i-1][j].class Solution {public:    int numDistinct(std::string S, std::string T) {        int n = S.size(),m = T.size();std::vector<std::vector<int>> dp(n+1,std::vector<int>(m+1,0));for (int i = 0; i < n+1; i++){dp[i][0] = 1;}for (int i = 1; i <= n; i++){for (int j = 1; j <= m; j++){if(S[i-1] == T[j-1]) dp[i][j] = dp[i-1][j-1] + dp[i-1][j];else dp[i][j] = dp[i-1][j];}}return dp[n][m];    }};

leetcode 是什東東有點不懂

裡面有很編程多面試的題目，可以線上編譯運行。難度比較高。如果自己能都做出來，對面大公司很有協助。我就是做的那裡的題目。
 
leetcode oj提交代碼方式是怎的？

不能寫main函數，你需要的是按照class Solution給的介面來實現它的一個成員函數

給一個參考答案
#include <sstream>using namespace std;class Solution {public: void reverseWords(string &s) { string ans = "", temp; stringstream sin(s); while(sin >> temp) { if(ans != "") { ans = temp + " " + ans; } else { ans = temp; } } s = ans; }};