描述

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

Insert a character
Delete a character
Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"

Output: 3

Explanation:

horse -> rorse (replace 'h' with 'r')

rorse -> rose (remove 'r')

rose -> ros (remove 'e')

Example 2:

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')

思路：动态规划

这是一个经典的动态规划问题，思路参考斯坦福的课程：http://www.stanford.edu/class/cs124/lec/med.pdf

这里把加2变成加1即可

dp[i][0] = i;
dp[0][j] = j;
dp[i][j] = dp[i - 1][j - 1], if word1[i - 1] = word2[j - 1];
dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i - 1][j] + 1, dp[i][j - 1] + 1), otherwise.

class Solution {

public:

    int minDistance(string word1, string word2) {

        int m = word1.size(), n = word2.size();

        vector<vector<int> > dp(m+, vector<int>(n+, ));

        for(int i = ;i<=m;++i)

            dp[i][] = i;

        for(int i = ;i<=n;++i)

            dp[][i] = i;

        for(int i = ;i<=m;++i){

            for(int j = ;j<=n;++j){

                if(word1[i-] == word2[j-])

                    dp[i][j] = dp[i-][j-];

                else

                    dp[i][j] = min(dp[i-][j-], min(dp[i][j-], dp[i-][j])) + ;

            }

        }

        return dp[m][n];

    }

};

巴特西

【LeetCode】【动态规划】Edit Distance

描述

思路：动态规划

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