LeetCode072 Edit Distance

来源：互联网发布：天下手游激活码淘宝编辑：程序博客网时间：2024/06/07 17:29

详细见：leetcode.com/problems/edit-distance

Java Solution: github

package leetcode;/** * 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 *//** * @author      zxwtry * @email       zxwtry@qq.com * @project     OJ * @package     leetcode * @file        P072_EditDistance.java * @type        P072_EditDistance * @date        2017年3月7日 下午2:45:17 * @details     Solution1: AC 16ms 35.42% * @details     Solution2: AC 12ms 77.30% * @details     Solution3: AC 17ms 24.20% */public class P072_EditDistance {static class Solution1 {    public int minDistance(String word1, String word2) {        if (word1 == null || word2 == null)             return minDistance(word1 == null ? "" : word1, word2 == null ? "" : word2);        if (word1.length() < word2.length())            return minDistance(word2, word1);        int[][] dp = new int[2][word2.length() + 1];        for (int j = 0; j <= word2.length(); j ++)            dp[0][j] = j;        for (int i = 1; i <= word1.length(); i ++) {            dp[1][0] = i;             for (int j = 1; j <= word2.length(); j ++) {                dp[1][j] = Math.min(Math.min(dp[0][j] + 1, dp[1][j - 1] + 1),                         dp[0][j - 1] + (word1.charAt(i - 1) == word2.charAt(j - 1) ? 0 : 1));            }            System.arraycopy(dp[1], 0, dp[0], 0, dp[0].length);        }        return dp[1][word2.length()];    }}static class Solution2 {public int minDistance(String word1, String word2) {if (word1 == null)return minDistance("", word2);if (word2 == null)return minDistance("", word1);int len1 = word1.length(), len2 = word2.length();if (len1 == 0 || len2 == 0)return len1 + len2;int[][] m = new int[len1 + 1][len2 + 1];for (int i = 0; i <= len1; i ++)m[i][0] = i;for (int j = 1; j <= len2; j ++)m[0][j] = j;for (int i = 1; i <= len1; i ++)for (int j = 1; j <= len2; j ++) {m[i][j] = Math.min(Math.min(m[i - 1][j] + 1, m[i][j - 1] + 1),         m[i - 1][j - 1] + (word1.charAt(i - 1) == word2.charAt(j - 1) ?  0 : 1));}return m[len1][len2];}}static class Solution3 {        public int minDistance(String word1, String word2) {            if (word1 == null || word2 == null)                 return minDistance(word1 == null ? "" : word1, word2 == null ? "" : word2);            if (word1.length() < word2.length())                return minDistance(word2, word1);            int[][] dp = new int[2][word2.length() + 1];            for (int j = 0; j <= word2.length(); j ++)                dp[0][j] = j;            for (int i = 1; i <= word1.length(); i ++) {                dp[1][0] = i;                 for (int j = 1; j <= word2.length(); j ++) {                    dp[1][j] = Math.min(Math.min(dp[0][j] + 1, dp[1][j - 1] + 1),                             dp[0][j - 1] + (word1.charAt(i - 1) == word2.charAt(j - 1) ? 0 : 1));                }                int[] tmp = dp[0];                dp[0] = dp[1];                dp[1] = tmp;            }            return dp[0][word2.length()];        }    }}

C Solution: github

/*    url: leetcode.com/problems/edit-distance    minDistance:  AC 9ms 23.81%    minDistance2: AC 9ms 23.81%*/#include <stdio.h>#include <stdlib.h>#include <string.h>int _min(int a, int b, int c) {    if (a < b) return  a < c ? a : c;    return b < c ? b : c;}int minDistance(char* w1, char* w2) {    int n1 = strlen(w1), n2 = strlen(w2), i = 0, j = 0;    int ** m = (int**) malloc(sizeof(int*) * (n1 + 1));    for (i = 0; i <= n1; i ++)        m[i] = (int*) malloc(sizeof(int) * (n2 + 1));    for (j = 0; j <= n2; j ++)        m[0][j] = j;    for (i = 0; i <= n1; i ++)        m[i][0] = i;    for (i = 0; i < n1; i ++) {        for (j = 0; j < n2; j ++) {            m[i+1][j+1] = _min(m[i][j]+(w1[i] == w2[j] ? 0 : 1),                 m[i+1][j] + 1, m[i][j+1] + 1);        }    }    j = m[n1][n2];    for (i = 0; i <= n1; i ++)        free(m[i]);    free(m);    return j;}void swap(int** m1, int** m2) {    int* t = *m1;    *m1 = *m2;    *m2 = t;}int minDistance2(char* w1, char* w2) {    int n1 = strlen(w1), n2 = strlen(w2), i = 0, j = 0;    int *m1 = (int*) malloc(sizeof(int) * (n2 + 1));    int *m2 = (int*) malloc(sizeof(int) * (n2 + 1));    for (j = 0; j <= n2; j ++)        m1[j] = j;    for (i = 0; i < n1; i ++) {        m2[0] = i+1;        for (j = 0; j < n2; j ++) {            m2[j+1] = _min(m1[j]+(w1[i] == w2[j] ? 0 : 1),                 m2[j] + 1, m1[j+1] + 1);        }        swap(&m1, &m2);    }    j = m1[n2];    free(m1);    free(m2);    return j;}int main() {    char* w1 = "b";    char* w2 = "";    printf("answer is %d\r\n", minDistance(w1, w2));    return 0;}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/edit-distance/    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年4月16日    @details:    Solution: 342ms 17.65%    @details:    Solution: 199ms 94.33%'''class Solution(object):    def minDistance(self, w1, w2):        """        :type w1: str        :type w2: str        :rtype: int        """        n1 = 0 if w1 == None else len(w1)        n2 = 0 if w2 == None else len(w2)        dp = [[0 for j in range(n2+1)] for i in range(n1+1)]        for j in range(n2+1): dp[0][j] = j        for i in range(n1+1): dp[i][0] = i        for i in range(n1):            for j in range(n2):                dp[i+1][j+1] = min(dp[i][j]+(0 if w1[i] == w2[j] else 1),                                   dp[i+1][j]+1, dp[i][j+1]+1)        return dp[n1][n2]class Solution2(object):    def minDistance(self, w1, w2):        """        :type w1: str        :type w2: str        :rtype: int        """        n1 = 0 if w1 == None else len(w1)        n2 = 0 if w2 == None else len(w2)        if n1 == 0 or n2 == 0: return n1+n2        d1, d2 = [0]*(n2+1), [0]*(n2+1)        for j in range(n2+1): d1[j] = j        for i in range(n1):            d2[0] = i+1            for j in range(n2):                d2[j+1] = min(d1[j]+(0 if w1[i] == w2[j] else 1),                                   d2[j]+1, d1[j+1]+1)            t = d1            d1 = d2            d2 = t        return d1[n2]    if __name__ == "__main__":    w1 = "ab"    w2 = "bc"    print(Solution().minDistance(w1, w2))

0 0