给定两个单词 word1 和 word2,计算出将 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')思路:dp来做,设 dp[i][j]为word1的前i个字符到word2的前j个字符的编辑距离。则状态转移方程为:
dp[i][j]=min(dp[i][j-1]+1, dp[i-1][j]+1, dp[i-1][j-1] (if(words1[i-1])==words2[j-1]), dp[i-1][j-1] +1(if(words1[i-1])!=words2[j-1]))。其中:
1、dp[i][j-1]+1,表示word2插入word2[j],或者word1删除word1[i]
2、dp[i-1][j]+1,表示word2删除word2[j],或者word1插入word1[i]
3、dp[i-1][j-1] (if(words1[i-1])==words2[j-1]),不动即可。
4、 dp[i-1][j-1] +1(if(words1[i-1])!=words2[j-1]),最后一个字符替换下。
边界条件,dp[i][0]=i, dp[0][j]=j,代码实现如下:
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, 0)); for (int i = 0; i <= m; ++i) dp[i][0] = i; for (int j = 1; j <= n; ++j) dp[0][j] = j; for (int i = 1; i <= m; ++i){ for (int j = 1; j <= n; ++j){ int a = (word1[i-1] == word2[j-1] ? dp[i - 1][j - 1] : dp[i - 1][j - 1] + 1); dp[i][j] = min(min(dp[i][j - 1] + 1, dp[i - 1][j] + 1), a); } } return dp[m][n]; } };