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  • [LeetCode] Edit Distance(很好的DP)

    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

    用m*n的矩阵vector中的(i,j)存储从word1中(0...i)到word2中(0...j)的最小变化次数

    矩阵中,从上到下对应删除操作,从左到右对应插入操作,从左上到右下对应Replace操作。

    class Solution {
    public:
        int minDistance(string word1, string word2) { //1->2
            if(word1==word2)
                return 0;
            int len1 = word1.size();
            int len2 = word2.size();
            vector<int> temp(len2+1,0);
            vector<vector<int> > vec(len1+1,temp);
    
            for(int i=1;i<=len1;i++){
                vec[i][0] = i;
            }
            for(int j=1;j<=len2;j++){
                vec[0][j] = j;
            }
            for(int i=1;i<=len1;i++){
                for(int j=1;j<=len2;j++){
                    int tem1,tem2;
    
                    tem1 = min(vec[i-1][j]+1,vec[i][j-1]+1);
                    tem2 = word1[i-1]==word2[j-1] ? vec[i-1][j-1]:vec[i-1][j-1]+1;
                    vec[i][j] = min(tem1,tem2);
                }//end for
            }//end for
            return vec[len1][len2];
        }
    };
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  • 原文地址:https://www.cnblogs.com/Xylophone/p/3919112.html
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