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【leetcode】Distinct Subsequences(hard)
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"
意思是指可以通过多少种方式删除S的一部分字母使得S变为T。
思路：真高兴，又做出来了~~
用ways[m][n]存储 S[0~m-1]变为T[0~n-1]的方式
那么 ways[i][j] = ways[i-1][j] //扔掉S[i-1]
+((S[i-1] == T[j-1]) ? ways[i-1][j-1] : 0); //当前S与T的字母匹配,则需加上S[0~m-2]变为T[0~n-2]的方式数
class Solution {public: int numDistinct(string S, string T) { int slen = S.length(); int tlen = T.length(); if(slen < tlen) return 0; vector<vector<int>> ways(slen + 1, vector<int>(tlen + 1, 0)); ways[0][0] = 1; for(int i = 1; i < slen + 1; i++) { ways[i][0] = 1; //若T没有字母那么只有一种方式令S变为T：删除S全部的字母 } for(int i = 1; i < slen + 1; i++) { for(int j = 1; j < tlen + 1; j++) { ways[i][j] = ways[i-1][j] //扔掉当前的 +((S[i-1] == T[j-1]) ? ways[i-1][j-1] : 0); //当前S与T的字母匹配 } } return ways[slen][tlen]; }};
看了看别人的答案，发现我们的代码几乎是一模一样，难道说题做多了大家的风格都一样了吗？
有个优化的方法，因为在计算ways[i][j]时，只用到了ways[i-1]行的信息，所以没有必要存储所有的历史信息，只要存上一行的就好。
/** * Further optimization could be made that we can use only 1D array instead of a * matrix, since we only need data from last time step. */int numDistinct(string S, string T) { int m = T.length(); int n = S.length(); if (m > n) return 0; // impossible for subsequence vector<int> path(m+1, 0); path[0] = 1; // initial condition for (int j = 1; j <= n; j++) { // traversing backwards so we are using path[i-1] from last time step for (int i = m; i >= 1; i--) { path[i] = path[i] + (T[i-1] == S[j-1] ? path[i-1] : 0); } } return path[m];}
【leetcode】Distinct Subsequences(hard)