[LeetCode] 72. Edit Distance 编辑距离

2021年09月15日 阅读数:2
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Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.html

You have the following 3 operations permitted on a word:python

  1. Insert a character
  2. Delete a character
  3. Replace a character

Example 1:app

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:post

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

给2个单词,求从一个单词变成另外一个单词须要的步骤,有三种变换方式,插入,删除和替换。ui

解法:dp, dp[i][j]表示从word1的前i个字符转换到word2的前j个字符所须要的步骤。url

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

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)blog

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++:rem

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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