c++ DP and Explanation


  • 0
    D

    for "sea" and "eat"
    current id i and j
    if word1[i] == word[j], then dp[i][j] = dp[i - 1][j - 1];
    else
    delete word1[i],then dp = dp[i - 1][j] + 1;
    delete word2[j],then dp = dp[i][j - 1] + 1;
    so dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;
    '''
    class Solution {
    public:
    int minDistance(string word1, string word2) {
    int len1 = word1.size();
    int len2 = word2.size();
    vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0));

        for (int i = 1; i <= len1; i++)
            dp[i][0] = i;
        for (int j = 1; j <= len2; j++)
            dp[0][j] = j;
        
        for (int i = 1; i <= len1; i++)
        {
            for (int j = 1; j <= len2; 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], dp[i][j - 1]) + 1;
            }
        }
        return dp[len1][len2];
    }
    

    };
    '''


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