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| 1 | +想法:定義 ways[i][j] 是 s[0 .. i - 1] 可以組成 t[0 .. j - 1] 的種類;那麼 |
| 2 | +ways[i][j] = ways[i - 1][j - 1](用到 s[i - 1] == t[j - 1]) + ways[i - 1][j](不使用最後元素) when s[i - 1] == t[j - 1] ; |
| 3 | +ways[i][j] = ways[i - 1][j] otherwise |
| 4 | +edgecase : ways[i][0] = appearance of t[0] in s[0 .. i - 1] (t[0] 的出現次數代表 s[0 ..i] 可組成 t[0] 的方式) |
| 5 | +因此我們只需依序填表即可,最後輸出 ways[n][m] |
| 6 | + |
| 7 | +Time Complexity : O(mn) for filling table and each grid takes O(1) time |
| 8 | +Space Complexity : O(mn) for the dp(ways) table |
| 9 | + |
| 10 | +class Solution { |
| 11 | +public: |
| 12 | + int numDistinct(string s, string t) { |
| 13 | + int n = s.length() , m = t.length() ; |
| 14 | + vector<vector<double>> ways(n + 1 , vector<double>(m + 1)) ; // both unsigned long long int and double won't overflow |
| 15 | + |
| 16 | + // initialize edge cases |
| 17 | + ways[1][1] = (s[0] == t[0])? 1 : 0 ; |
| 18 | + for(int i = 2 ; i <= n ; i++) |
| 19 | + ways[i][1] = ways[i - 1][1] + ((s[i - 1] == t[0])? 1 : 0) ; |
| 20 | + |
| 21 | + |
| 22 | + for(int i = 1 ; i <= n ; i++) { |
| 23 | + // now consider s[i - 1] letter |
| 24 | + for(int j = 2 ; j <= min(i , m) ; j++) { |
| 25 | + // ways[i][j] means the ways that s[0 .. i-1] can make t[0 .. j-1] |
| 26 | + if ( s[i - 1] == t[j - 1] ) { |
| 27 | + ways[i][j] = ways[i - 1][j - 1] + ways[i - 1][j]; |
| 28 | + } |
| 29 | + else |
| 30 | + ways[i][j] = ways[i - 1][j] ; |
| 31 | + } |
| 32 | + } |
| 33 | + |
| 34 | + |
| 35 | + return ways[n][m] ; |
| 36 | + } |
| 37 | +}; |
| 38 | + |
| 39 | +// 由於我們先前的填表其實僅用到上一列的數據,因此我們可以僅儲存一列數據即可(空間優化) |
| 40 | + |
| 41 | +Time Complexity : O(n) for storing only one row record |
| 42 | + |
| 43 | +class Solution { |
| 44 | +public: |
| 45 | + int numDistinct(string s, string t) { |
| 46 | + int n = s.length() , m = t.length() ; |
| 47 | + vector<double> ways(m + 1) ; |
| 48 | + |
| 49 | + // initialize edge cases |
| 50 | + ways[1] = (s[0] == t[0])? 1 : 0 ; |
| 51 | + |
| 52 | + for(int i = 2 ; i <= n ; i++) { |
| 53 | + // now consider s[i - 1] letter |
| 54 | + cout << "1 : " << ways[1] << endl ; |
| 55 | + for(int j = min(i , m) ; j > 1 ; j--) { |
| 56 | + if ( s[i - 1] == t[j - 1] ) { |
| 57 | + ways[j] += ways[j - 1] ; |
| 58 | + } |
| 59 | + } |
| 60 | + if ( s[i - 1] == t[0] ) |
| 61 | + ways[1]++ ; |
| 62 | + } |
| 63 | + |
| 64 | + |
| 65 | + return ways[m] ; |
| 66 | + } |
| 67 | +}; |
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