leetcode@ [72/115] Edit Distance & Distinct Subsequences (Dynamic Programming)

https://leetcode.com/problems/edit-distance/

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

用dp[i][j] 来表示 长度为 i 的 word1 经过 dp[i][j]次变换 可以得到长度为 j 的word2，那么我们主要考察两种情况，第一种是：word1[i] == word2[j]，那么这个问题的规模便转换成了：dp[i][j] = dp[i-1][j-1]. 第二种情况是：word1[i] != word2[j]，那么我们可以删除掉word1中的第 i 个字符，或者我们可以把 word1中的第 i 个字符换成与 word2[j] 相同的字符，或者我们还可以同时 删除 word1[i] 和 word2[j]. 于是状态转移方程为：dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1.

class Solution {
public:
int numDistinct(string s, string t) {
int m = s.length(), n = t.length();
vector<vector<int> > dp(m+1, vector<int>(n+1, 0));

for(int i=0;i<=m;++i) dp[i][0] = 1;

for(int i=1;i<=m;++i) {
for(int j=1;j<=n;++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[m]
;
}
};

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