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    leetcode的题目

    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

    这是一个动态规划的题目

    wiki上的递推公式如下

    代码如下

     1   int minDistance(string word1, string word2) {
     2         int **dp = new int*[word1.length() + 1];
     3         for(int i = 0; i < word1.length() + 1; i ++)
     4             dp[i] = new int[word2.length() + 1];
     5 
     6         for(int i = 0; i < word1.length() + 1; i ++)
     7             dp[i][0] = i;
     8         for(int i = 0; i < word2.length() + 1; i ++)
     9             dp[0][i] = i;
    10         
    11         for(int i = 1; i < word1.length() + 1; i ++)
    12             for(int j = 1; j < word2.length() + 1; j ++)
    13             {
    14                 dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;//insert or delete
    15 
    16                 if(word1[i - 1] != word2[j - 1])//modify
    17                 {
    18                     dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + 1);
    19                 }
    20                 else
    21                 {
    22                     dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
    23                 }
    24             }
    25 
    26             return dp[word1.length()][word2.length()];
    27     }

     时间复杂度和空间复杂度都是O(M*N)

    如果不要求计算修改过程, 那么空间复杂度可以优化到线性

    感觉自己对动态规划还是很不敏感

    查阅相关资料后发现, 这个题目属于文本比较算法, 还有很多相关的内容, 比如说LCS和一些其他更加高级的算法, 以后有空再研究一下

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  • 原文地址:https://www.cnblogs.com/stevenczp/p/3928087.html
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