Edit Distance
最后更新于:2022-04-01 22:57:13
**一. 题目描述**
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:
Insert a character
Delete a character
Replace a character
**二. 题目分析**
给定两个字符串word1和word2,算出讲word1转化成word2所需的最少编辑操作次数。允许的编辑操作包括以下三种:
将一个字符替换成另一个字符
在字符串中插入一个字符
删除字符串中的一个字符
例如将A(abc)转成B(acbc):
你可选择以下操作:
acc (b→c)替换
acb (c→b)替换
acbc (→c)插入
这不是最少编辑次数,因为其实只需要删除第二个字符`c`就可以了,这样只需操作一次。
使用`i`表示字符串`word1`的下标(从下标1开始),使用`j`表示字符串`word2`的下标。 用`k[i][j]`来表示`word1[1, ... , i]`到`word2[1, ... , j]`之间的最少编辑操作数。则有以下规律:
~~~
k[i][0] = i;
k[0][j] = j;
k[i][j] = k[i - 1][j - 1] (if word1[i] == word2[j])
k[i][j] = min(k[i - 1][j - 1],
k[i][j - 1],
k[i - 1][j]) + 1 (if word1[i] != word2[j])
~~~
**三. 示例代码**
~~~
#include
#include
#include
using namespace std;
class Solution
{
public:
int minDistance(const string &word1, const string &word2)
{
const size_t m = word1.size() + 1;
const size_t n = word2.size() + 1;
vector > k(m, vector(n));
for (size_t i = 0; i < m; ++i)
k[i][0] = i;
for (size_t j = 0; j < n; ++j)
k[0][j] = j;
for (size_t i = 1; i < m; ++i)
{
for (size_t j = 1; j < n; ++j)
{
if (word1[i - 1] == word2[j - 1])
k[i][j] = k[i - 1][j - 1];
else
k[i][j] = min(k[i - 1][j - 1], min(k[i - 1][j], k[i][j - 1])) + 1;
}
}
return k[m - 1][n - 1];
}
};
~~~
![](https://docs.gechiui.com/gc-content/uploads/sites/kancloud/2016-01-05_568bb5f0b9ec2.jpg)
**四. 小结**
动态规划的经典题目,要快速写出状态转移方程还是有点难度的。
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