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:
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’)
每个位置都有四种状态:匹配,删除,插入,修改。
匹配:找前一段编辑距离就好。
否则,取下列三个操作中编辑距离最短的:
class Solution {
public int minDistance(String word1, String word2) {
if(word1.isEmpty()) return word2.length();
if(word2.isEmpty()) return word1.length();
int n = word1.length();
int m = word2.length();
if(word1.charAt(n - 1) == word2.charAt(m - 1)) {
return minDistance(word1.substring(0, n - 1), word2.substring(0, m - 1));
} else {
int delete = 1 + minDistance(word1.substring(0, n - 1), word2.substring(0, m));
int insert = 1 + minDistance(word1.substring(0, n), word2.substring(0, m - 1));
int update = 1 + minDistance(word1.substring(0, n - 1), word2.substring(0, m - 1));
return Math.min(Math.min(delete, insert), update);
}
}
}
备忘录法复杂度:O(nm) 时间,O(nm) 空间。
class Solution {
private Integer[][] memo;
public int minDistance(String word1, String word2) {
int n = word1.length();
int m = word2.length();
memo = new Integer[n + 1][m + 1];
return minDistance(word1.toCharArray(), n, word2.toCharArray(), m);
}
private int minDistance(char[] w1, int i, char[] w2, int j) {
if(memo[i][j] != null) return memo[i][j];
int ret = 0;
if(i == 0) {
ret = j;
} else if(j == 0) {
ret = i;
} else {
if(w1[i - 1] == w2[j - 1]) {
ret = minDistance(w1, i - 1, w2, j - 1);
} else {
int delete = 1 + minDistance(w1, i - 1, w2, j);
int insert = 1 + minDistance(w1, i, w2, j - 1);
int update = 1 + minDistance(w1, i - 1, w2, j - 1);
ret = Math.min(Math.min(delete, insert), update);
}
}
memo[i][j] = ret;
return ret;
}
}
状态定义:dp[i][j] 表示 word1[0…i -1] 与 word2[0…j -1] 的编辑距离。
状态转移:
初始化:
返回值:dp[n][m]
复杂度:O(nm) 时间,O(nm) 空间。(tips:可以尝试优化成O(n + m) 的空间复杂度)
class Solution {
public int minDistance(String word1, String word2) {
int n = word1.length();
int m = word2.length();
int[][] dp = new int[n + 1][m + 1];
for(int i = 0; i <= n; i ++) {
dp[i][0] = i;
}
for(int i = 0; i <= m; i ++) {
dp[0][i] = i;
}
for(int i = 1; i <= n; i ++) {
for(int j = 1; j <= m; j ++) {
if(word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
}
}
}
return dp[n][m];
}
}