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编辑距离

目前感受下来,这种字符串计算的,最好都多预留一个,比较好算。
int minDistance(string word1, string word2) {
    int m = word1.size();
    int n = word2.size();

    vector<vector<int>> dp(m+1, vector<int>(n+1, 0));

    // 初始化
    for (int i = 1; i <= m; i++) dp[i][0] = i;
    for (int j = 1; j <= n; j++) dp[0][j] = j;

    // 计算 dp 数组
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (word1[i-1] == word2[j-1]) {
                dp[i][j] = dp[i-1][j-1];
            } else {
                int m1 = min(dp[i][j-1], dp[i-1][j]);
                dp[i][j] = min(m1, dp[i-1][j-1]) + 1;
            }
        }
    }

    return dp[m][n];
}
 
你觉得这篇文章怎么样?
YYDS
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