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

思路：普通的DP很好写，问题是路径压缩的DP压缩如果进行，空间如何重复利用，和背包问题有点像，留待第二遍再解决

class Solution {
public:
    int minDistance(string word1, string word2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int ** dp = new int*[word1.size() + 1];
        for(int i = 0; i < word1.size() + 1; i++){
            dp[i] = new int[word2.size() + 1];
        }

        for(int i = 0; i < word1.size() + 1; i++){
            dp[i][0] = i;
        }
        for(int j = 0; j < word2.size() + 1; j++){
            dp[0][j] = j;
        }

        for(int i = 1; i < word1.size() + 1; i++){
            for(int j = 1; j < word2.size() + 1; j++){
                if (word1[i-1] == word2[j-1]){
                    //相等的情况下，最后一个元素不占用操作
                    dp[i][j] = dp[i-1][j-1];
                }else{
                    //delete一种串
                    dp[i][j] = min(dp[i-1][j] + 1,dp[i][j-1] + 1);
                    //i和j的位置进行replace
                    dp[i][j] = min(dp[i][j],dp[i-1][j-1] + 1);
                }
            }
        }    

        int ans = dp[word1.size()][word2.size()];
        for(int i = 0; i < word1.size() + 1; i++){
            delete []dp[i];
        }
        delete []dp;
        return ans;
    }
};

原文链接: https://www.cnblogs.com/kwill/p/3182882.html

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