Leetcode题解：十几行代码计算编辑距离

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/jining11/article/details/83624520

题目要求

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

求解思路

这是一个经典的动态规划问题了，以前做DNA序列比较时也完成过。

假设：

$字符串S_1和S_2分别长为l_1和l_2$
$E_{ij}代表S_1的前i个字符组成的串到S_2的前j个字符组成的串的最短编辑距离$

状态转移方程：
$E_{ij} =min(E_{i(j-1)}, E_{(i-1)j}, E_{(i-1)(j-1)}+diff(S_{1i}, S_{2j}))$
$diff(x, y)= \begin{cases} 0& \text{x=y}\\ 1& \text{x!=y} \end{cases}$
状态初始化
$E_{0j} = j, j = 0...l_2$
$E_{i0} = i, i = 0...l_1$

源代码

有两点需要注意，一个是字符串的下标需要-1，另一个是diff的实现方式，如果写成我一开始提交的版本即int(word1[i-1]!=word2[j-1])，执行速度会非常之慢，还是写成(word1[i-1]!=word2[j-1] ? 1 : 0)为好

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

满意了满意了