LeetCode 72 — Edit Distance(C++ Python)

题目:https://oj.leetcode.com/problems/edit-distance/

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

分析:

动态规划。设f[i][j]表示 word1[0,i] 和 word2[0,j] 之间的最小编辑距离。如果word2[j - 1] == word1[i - 1],则f[i][j] = f[i - 1][j - 1],否则选择3种操作中编辑距离最小的再加上1,具体见代码。(f[i-1][j-1]对应替换word1[i - 1]为word2[j - 1], f[i-1][j]对应删除word1[i - 1],f[i][j-1]对应在word1[i - 1]后面添加word2[j - 1]

C++实现:

class Solution {
public:
    int minDistance(string word1, string word2) {
		const int len1 = word1.size();
		const int len2 = word2.size();
		vector > f(len1 + 1, vector(len2 + 1));

		for(int i = 0; i <= len1; ++i)
		{
			f[i][0] = i;
		}
		for(int j = 0; j <= len2; ++j)
		{
			f[0][j] = j;
		}

		for(int i = 1; i <= len1; ++i)
			for(int j = 1; j <= len2; ++j)
			{
				if(word2[j - 1] == word1[i - 1]) 
				{
					f[i][j] = f[i - 1][j - 1];
				}
				else
				{
					int tmp = std::min(f[i][j - 1], f[i - 1][j]);
					f[i][j] = std::min(f[i - 1][j - 1], tmp) + 1;
				}
			}

		return f[len1][len2];
    }
};

Python实现:

class Solution:
    # @return an integer
    def minDistance(self, word1, word2):
        len1 = len(word1)
        len2 = len(word2)
        f = [[0] * (len2 + 1) for i in range(len1 + 1)] 
        		
        for i in range(len1 + 1):
            f[i][0] = i
        for j in range(len2 + 1):
            f[0][j] = j

        for i in range(1, len1 + 1):
            for j in range(1, len2 + 1):
                if word2[j - 1] == word1[i - 1]:
                    f[i][j] = f[i - 1][j - 1]
                else:
                    f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1
                    	
        return f[len1][len2]

        感谢阅读!

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