[LeetCode] 72. Edit Distance 编辑距离

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

给2个单词，求从一个单词变成另一个单词需要的步骤，有三种变换方式，插入，删除和替换。

解法：DP

Python:  Time: O(n * m)  Space: O(n + m)

class Solution:
    # @return an integer
    def minDistance(self, word1, word2):
        if len(word1) < len(word2):
            return self.minDistance(word2, word1)
        
        distance = [i for i in xrange(len(word2) + 1)]
        
        for i in xrange(1, len(word1) + 1):
            pre_distance_i_j = distance[0]
            distance[0] = i
            for j in xrange(1, len(word2) + 1):
                insert = distance[j - 1] + 1
                delete = distance[j] + 1
                replace = pre_distance_i_j
                if word1[i - 1] != word2[j - 1]:
                    replace += 1
                pre_distance_i_j = distance[j]
                distance[j] = min(insert, delete, replace)

        return distance[-1]

Python:  Time: O(n * m)  Space: O(n * m)

class Solution:
    # @return an integer
    def minDistance(self, word1, word2):        
        distance = [[i] for i in xrange(len(word1) + 1)]
        distance[0] = [j for j in xrange(len(word2) + 1)]
        
        for i in xrange(1, len(word1) + 1):
            for j in xrange(1, len(word2) + 1):
                insert = distance[i][j - 1] + 1
                delete = distance[i - 1][j] + 1
                replace = distance[i - 1][j - 1]
                if word1[i - 1] != word2[j - 1]:
                    replace += 1
                distance[i].append(min(insert, delete, replace))
                
        return distance[-1][-1]

C++：

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1 = word1.size(), n2 = word2.size();
        int dp[n1 + 1][n2 + 1];
        for (int i = 0; i <= n1; ++i) dp[i][0] = i;
        for (int i = 0; i <= n2; ++i) dp[0][i] = i;
        for (int i = 1; i <= n1; ++i) {
            for (int j = 1; j <= n2; ++j) {
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        return dp[n1][n2];
    }
};

　

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