Edit Distance 動態規劃

Edit Distance

思路來源：https://oj.leetcode.com/discuss/10426/my-o-mn-time-and-o-n-space-solution-using-dp-with-explanation

My O(mn) time and O(n) space solution using DP with explanation

Use f[i][j] to represent the shortest edit distance between word1[0,i) and word2[0, j). Then compare the last character of word1[0,i) and word2[0,j), which are c and d respectively (c == word1[i-1], d == word2[j-1]):

if c == d, then : f[i][j] = f[i-1][j-1]

Otherwise we can use three operations to convert word1 to word2:

(a) if we replaced c with d: f[i][j] = f[i-1][j-1] + 1;

(b) if we added d after c: f[i][j] = f[i][j-1] + 1;

(c) if we deleted c: f[i][j] = f[i-1][j] + 1;

Note that f[i][j] only depends on f[i-1][j-1], f[i-1][j] and f[i][j-1], therefore we can reduce the space to O(n) by using only the (i-1)th array and previous updated element(f[i][j-1]).