class Solution {
public:
int minDistance(string word1, string word2) {
}
};
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：
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 <= 500word1 和 word2 由小写英文字母组成原站题解
python3 解法, 执行用时: 92 ms, 内存消耗: 17.8 MB, 提交时间: 2023-04-19 10:22:51
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
import functools
@functools.lru_cache(None)
def helper(i, j):
if i == len(word1) or j == len(word2):
return len(word1) - i + len(word2) - j
if word1[i] == word2[j]:
return helper(i + 1, j + 1)
else:
inserted = helper(i, j + 1)
deleted = helper(i + 1, j)
replaced = helper(i + 1, j + 1)
return min(inserted, deleted, replaced) + 1
return helper(0, 0)
java 解法, 执行用时: 5 ms, 内存消耗: 41.4 MB, 提交时间: 2023-04-19 10:22:07
class Solution {
public int minDistance(String word1, String word2) {
int n = word1.length();
int m = word2.length();
// 有一个字符串为空串
if (n * m == 0) {
return n + m;
}
// DP 数组
int[][] D = new int[n + 1][m + 1];
// 边界状态初始化
for (int i = 0; i < n + 1; i++) {
D[i][0] = i;
}
for (int j = 0; j < m + 1; j++) {
D[0][j] = j;
}
// 计算所有 DP 值
for (int i = 1; i < n + 1; i++) {
for (int j = 1; j < m + 1; j++) {
int left = D[i - 1][j] + 1;
int down = D[i][j - 1] + 1;
int left_down = D[i - 1][j - 1];
if (word1.charAt(i - 1) != word2.charAt(j - 1)) {
left_down += 1;
}
D[i][j] = Math.min(left, Math.min(down, left_down));
}
}
return D[n][m];
}
}
python3 解法, 执行用时: 140 ms, 内存消耗: 18.8 MB, 提交时间: 2023-04-19 10:21:47
'''
dp[i][j] word1的前i个字符和word2的前j个字符的编辑距离
'''
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
n = len(word1)
m = len(word2)
# 有一个字符串为空串
if n * m == 0:
return n + m
# DP 数组
D = [ [0] * (m + 1) for _ in range(n + 1)]
# 边界状态初始化
for i in range(n + 1):
D[i][0] = i
for j in range(m + 1):
D[0][j] = j
# 计算所有 DP 值
for i in range(1, n + 1):
for j in range(1, m + 1):
left = D[i - 1][j] + 1 # 删除一个字符
down = D[i][j - 1] + 1 # 插入一个字符
left_down = D[i - 1][j - 1] # 替换一个字符
if word1[i - 1] != word2[j - 1]:
left_down += 1
D[i][j] = min(left, down, left_down)
return D[n][m]