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
Example 1:
Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')别人的代码
Runtime: 6 ms, faster than 75.50% of Java online submissions for Edit Distance.
Memory Usage: 34.6 MB, less than 93.51% of Java online submissions for Edit Distance.
public class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length();
int n = word2.length();
int[][] cost = new int[m + 1][n + 1];
for(int i = 0; i <= m; i++) // 处理边界
cost[i][0] = i;
for(int i = 1; i <= n; i++)
cost[0][i] = i;
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(word1.charAt(i) == word2.charAt(j))
cost[i + 1][j + 1] = cost[i][j];
else {
int a = cost[i][j];
int b = cost[i][j + 1];
int c = cost[i + 1][j];
cost[i + 1][j + 1] = a < b ? (a < c ? a : c) : (b < c ? b : c);
cost[i + 1][j + 1]++;
}
}
}
return cost[m][n];
}
}定义：f(i, j) 为将word1的前i个字符转换成word2的前j个字符的最小cost（steps）
Case 1: word1[i] == word2[j]；则有f(i, j) = f(i - 1, j - 1)
Case 2: word1[i] != word2[j], then we must either insert, delete or replace, whichever is cheaper
f(i, j) = 1 + min { f(i, j - 1), f(i - 1, j), f(i - 1, j - 1) }
- f(i, j - 1) represents insert operation
- f(i - 1, j) represents delete operation
- f(i - 1, j - 1) represents replace operation
比如：
f(i, j - 1)represents insert operation，就表示在word1的第i个字符后插入一个与word2第j个字符相同的字符
f(i - 1, j) represents delete operation，就表示将word1的第i个字符删除
f(i - 1, j - 1) represents replace operation，就代表将word1的第i个字符替换成与word2第j个字符相同的字符
Above equations become the recursive definitions for DP.
Base Case:
f(0, k) = f(k, 0) = k
递归版本：
容易理解的递归算法（提交结果超时）：
class Solution {
public:
int minDistance(string word1, string word2) {
if(word1 == word2) return 0;
int m = word1.size();
int n = word2.size();
if(word1 == "") return n;
if(word2 == "") return m;
if(word1[0] == word2[0])
return minDistance(word1.substr(1), word2.substr(1));
else {
return min(minDistance(word1, word2.substr(1))+1, min(minDistance(word1.substr(1), word2)+1, minDistance(word1.substr(1), word2.substr(1))+1)); //递归调用，报错TLE
}
}
};