leetcode 72. Edit Distance

最新推荐文章于 2024-06-25 14:15:48 发布

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最新推荐文章于 2024-06-25 14:15:48 发布

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本文介绍了一个经典的字符串操作问题——编辑距离。通过动态规划的方法解决如何将一个字符串转换为另一个字符串的问题，提供了详细的算法实现，并解释了插入、删除和替换字符的操作。

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72. 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

这种题就是动态规划来做！f[i][j]代表word1的前i位变成word2的前j位需要多少步骤。

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