72. 编辑距离
给你两个单词 word1 和 word2, 请返回将 word1 转换成 word2 所使用的最少操作数 。
你可以对一个单词进行如下三种操作:
插入一个字符
删除一个字符
替换一个字符
示例 1:
输入:word1 = "horse", word2 = "ros"
输出:3
解释:
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')
示例 2:
输入:word1 = "intention", word2 = "execution"
输出:5
解释:
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')
提示:
0 <= word1.length, word2.length <= 500
word1 和 word2 由小写英文字母组成
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/edit-distance
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
解题思路:
class Solution {
public:
int minDistance(string word1, string word2)
{
int m = word1.size(), n = word2.size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
for (int i = 1; i <= n; ++i)
dp[0][i] = i;
for (int i = 1; i <= m; ++i)
dp[i][0] = i;
for (int i = 1; i <= m; ++i)
{
for (int j = 1; j <= n; ++j)
{
int left = dp[i][j - 1] + 1;
int up = dp[i - 1][j] + 1;
int mid;
if (word1[i - 1] == word2[j - 1])
mid = dp[i - 1][j - 1];
else
mid = dp[i - 1][j - 1] + 1;
dp[i][j] = min(min(left, up), mid);
}
}
return dp[m][n];
}
};