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| 1 | +#include <stdio.h> |
| 2 | +#include <stdlib.h> |
| 3 | +#include <string.h> |
| 4 | + |
| 5 | +int min(int a, int b, int c) { |
| 6 | + int min = a; |
| 7 | + if (b < min) min = b; |
| 8 | + if (c < min) min = c; |
| 9 | + return min; |
| 10 | +} |
| 11 | + |
| 12 | +int minDistance(char *word1, char *word2) { |
| 13 | + int len1 = strlen(word1); |
| 14 | + int len2 = strlen(word2); |
| 15 | + |
| 16 | + /* these two lines can be commented out */ |
| 17 | + if (len1 == 0) return len2; |
| 18 | + if (len2 == 0) return len1; |
| 19 | + |
| 20 | + int(*d)[len2 + 1] = (int (*)[len2 + 1])calloc((len1 + 1) * (len2 + 1), sizeof(int)); |
| 21 | + int i, j; |
| 22 | + |
| 23 | + d[0][0] = 0; /* dummy */ |
| 24 | + for (i = 1; i <= len1; i++) d[i][0] = i; |
| 25 | + for (j = 1; j <= len2; j++) d[0][j] = j; |
| 26 | + |
| 27 | + for (i = 1; i <= len1; i++){ |
| 28 | + for (j = 1; j <= len2; j++) { |
| 29 | + /* d[i][j] represents distance of word1[0..i-1] and word2[0..j-1] */ |
| 30 | + if (word1[i - 1] == word2[j - 1]) { |
| 31 | + d[i][j] = d[i - 1][j - 1]; |
| 32 | + } |
| 33 | + else { |
| 34 | + d[i][j] = min( |
| 35 | + d[i][j - 1] + 1, /* insert a character(word2[j-1]) to the tail of word1[0..i-1] */ |
| 36 | + d[i - 1][j] + 1, /* delete a character(word1[i-1]) from the tail of word1[0..i-1] */ |
| 37 | + d[i - 1][j - 1] + 1 /* replace a character in tail of word1[0..i-1] */ |
| 38 | + ); |
| 39 | + } |
| 40 | + } |
| 41 | + } |
| 42 | + return d[len1][len2]; |
| 43 | +} |
| 44 | + |
| 45 | +int main() { |
| 46 | + printf("%d %d\n", minDistance("word", "wood"), 1); |
| 47 | + printf("%d %d\n", minDistance("word", "woord"), 1); |
| 48 | + printf("%d %d\n", minDistance("d", "word"), 3); |
| 49 | + printf("%d %d\n", minDistance("", "word"), 4); |
| 50 | + printf("%d %d\n", minDistance("abcd", "word"), 3); |
| 51 | + printf("%d %d\n", minDistance("ffff", "word"), 4); |
| 52 | + return 0; |
| 53 | +} |
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