Leetcode dp Edit Distance

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Edit Distance Total Accepted: 14997 Total Submissions: 59129My Submissions

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

題意：有兩個單詞，可以選擇以下三種操作：增、刪、替換一個字元，問需要經過多少次操作才能把第一個單詞變為第二個
思路：dp
f[i][j] 表示word1[1..j]到word2[1..i]的最小編輯距離
if word2[i] == word1[j], f[i][j] = f[i - 1][j - 1]
else f[i][j] = min{f[i - 1][j - 1], f[i][j - 1], f[i - 1][j]} + 1
複雜度：時間O(n)，空間O(1)

int minDistance(string word1, string word2) {if(word1.empty()) return word2.size();if(word2.empty()) return word1.size();vector<vector<int> > dp(word1.size() + 1, vector<int>(word2.size() + 1, 0)); ////初始化for(int i = 0; i <= word1.size(); ++i) dp[i][0] = i;for(int j = 0; j <= word2.size(); ++j) dp[0][j] = j;//迭代for(int i = 0; i < word1.size(); ++i){for(int j = 0; j < word2.size(); ++j){int ii  = i + 1, jj = j + 1;if(word1[i] == word2[j]) dp[ii][jj] = dp[ii - 1][jj - 1];else dp[ii][jj] = min(min(dp[ii][jj - 1], dp[ii - 1][jj]), dp[ii - 1][jj - 1]) + 1;}}return dp[word1.size()][word2.size()];}

Leetcode dp Edit Distance