day56|● 583. 两个字符串的删除操作 ● 72. 编辑距离

583. 两个字符串的删除操作

方法一: 想到和1143.最长公共子序列问题基本相同,求出最长公共子序列,除了最长公共子序列之外的字符都是必须删除的:用word1和word2 的长度减去公共子序列的长度。

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector> dp(word1.size()+1,vector(word2.size()+1,0));
        for(int i=1;i<=word1.size();i++){
            for(int j=1;j<=word2.size();j++){
                if(word1[i-1] == word2[j-1])  dp[i][j] = dp[i-1][j-1]+1;
                else dp[i][j] = max(dp[i-1][j] ,dp[i][j-1]);
            }
        }
        return word1.size()+word2.size()-2*dp[word1.size()][word2.size()];
    }
};

方法二:按照题意 写代码。dp[i][j] 代表使得下标0到i-1的word1和下标0到j-1的word2相同,所需要的最小步数。递推公式:判断 word1[i - 1]  word2[j - 1]是否相同,如果相同则不需要进行改变,dp[i][j] = dp[i - 1][j - 1];如果不同,则需要在word1 word2或者二者同时删除 中 取最小步数,

dp[i][j] = min({dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i-1][j - 1] + 2});

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector> dp(word1.size() + 1, vector(word2.size() + 1));
        for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
        for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
        for (int i = 1; i <= word1.size(); i++) {
            for (int j = 1; j <= word2.size(); j++) {
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = min({dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i-1][j - 1] + 2});
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

72. 编辑距离

同上,只是在不同时进行的操作不同。

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector> dp(word1.size()+1,vector(word2.size()+1,0));
        for(int i=0;i<=word1.size();i++)  dp[i][0] = i;
        for(int j=0;j<=word2.size();j++)  dp[0][j] = j;
        
        for(int i=1;i<=word1.size();i++){
            for(int j=1;j<=word2.size();j++){
                if(word1[i-1] == word2[j-1]){
                    dp[i][j] = dp[i-1][j-1];
                }else{
                    dp[i][j] = min({dp[i-1][j-1],dp[i-1][j],dp[i][j-1]})+1;
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

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