• leetcode 72. Edit Distance


    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')
    

    题意:一个字符串,插入,删除,或替换变道另一个字符串,输出最小操作数。

    思路:dp.
    可以抽象成一个矩阵。
    比如下面矩阵表示,比如第1行第3列,表示由空变到ab需要的操作数。第3行第1列表示由ac变到空需要的操作数。

      空 a b c
    空 0 1 2 3
    a  1 0 ?
    c  2
    c  3
    

    dp[i][j]表示,字符串s的前i个字母变到字符串t的前j个字母需要的最少的操作数。那么dp[i][j] = min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1]) + 1.
    分别对应替换,删除,添加。
    比如问好位置,可由空->aa替换后一个a,或者a->a添加一个b,或则空->aab删除个a.

    class Solution {
    public:
        int minDistance(string word1, string word2) {
            int n = word1.size();
            int m = word2.size();
            vector<vector<int> > dp(n+1, vector<int>(m+1));
            for (int i = 0; i <= n; ++i) {
                dp[i][0] = i;
            }
            for (int j = 0; j <= m; ++j) {
                dp[0][j] = j;
            }
            for (int i = 1; i <= n; ++i) {
                for (int j = 1; j <= m; ++j) {
                    if (word1[i-1] == word2[j-1]) {
                        dp[i][j] = dp[i-1][j-1];
                    }
                    else {
                        dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]),dp[i-1][j-1])+1;
                    }
                }
            }
            return dp[n][m];
        }
    };
    
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  • 原文地址:https://www.cnblogs.com/pk28/p/9573343.html
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