[Leetcode] Edit Distance

Edit Distance

Given 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

Problem Solving Ideas:

This problem calculates the editing distance of two texts. With the idea of dynamic planning. D[I][J] Represents the minimum number of transitions from string A (0..I) to B (0...J). Then there are:

Class Solution {public:    int mindistance (string word1, String word2) {        int m = Word1.length ();        int n = word2.length ();        if (m = = 0 | | n = = 0) {            return Max (M, n);        }        Vector<vector<int>> D (m + 1, vector<int> (n + 1, 0));        for (int i = 0; i<=n; i++) {            d[0][i] = i;        }        for (int i=0; i<=m; i++) {            d[i][0] = i;        }        for (int i = 1; i<=m; i++) {for            (int j = 1; j<=n; j + +) {                if (word1[i-1] = = Word2[j-1]) {                    D[i][j] = d[i-1][ J-1];                } else{                    D[i][j] = min (min (d[i][j-1], d[i-1][j]), d[i-1][j-1]) + 1;}}        }        return d[m][n];}    ;

The time complexity and space complexity of the above code are all O (n^2). You can change the code to the space complexity of O (n).

Class Solution {public:    int mindistance (string word1, String word2) {        int m = Word1.length ();        int n = word2.length ();        if (m = = 0 | | n = = 0) {            return Max (M, n);        }        Vector<int> d (n + 1, 0);        for (int i = 0; i<=n; i++) {            d[i] = i;        }        for (int i = 1; i<=m; i++) {            int left = i;            int leftup = i-1;            for (int j = 1; j<=n; j + +) {                int up = d[j];                int newval;                if (word1[i-1] = = Word2[j-1]) {                    newval = Leftup;                } else{                    newval = min (min (left, up), leftup) + 1;                }                Leftup = D[j];                D[J] = newval;                left = D[j];            }        }        return d[n];}    ;

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[Leetcode] Edit Distance