leetcode 72. Edit Distance

参考：/article/1532233.html

题意

求word1变成word2的最短编辑距离。（插入、删除、修改都记为一步操作）

题解

使用动态规划解决。dp[i][j] 表示word1子串（0~i-1）变成word2子串（0~j-1）所需的最短编辑距离。

状态转移方程为：

dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j] + 1, dp[i][j - 1] + 1) 　　　if(word1[i - 1] == word2[j - 1])

dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i - 1][j] + 1, dp[i][j - 1] + 1)　　if(word1[i - 1] != word2[j - 1])

代码

class Solution {
public:
int minDistance(string word1, string word2) {
int len1 = word1.length(), len2 = word2.length();
if(len2 == 0 || len1 == 0)
return max(len1, len2);
int dp[len1 + 1][len2 + 1];

for(int i = 0; i <= len1; i++)
dp[i][0] = i;
for(int j = 0; j <= len2; j++)
dp[0][j] = j;
for(int i = 1; i <= len1; i++)
{
for(int j = 1; j <= len2; j++)
{
int curDis;
if(word1[i - 1] == word2[j - 1])
curDis = dp[i - 1][j - 1];
else
curDis = dp[i - 1][j - 1] + 1;
curDis = min(curDis, dp[i - 1][j] + 1); // word1 delete word1[i]
curDis = min(curDis, dp[i][j - 1] + 1); // word1 insert word2[j]
dp[i][j] = curDis;
}
}
return dp[len1][len2];
}
};