给定两个单词 word1
和 word2
,返回使得 word1
和 word2
相同所需的最小步数。
每步 可以删除任意一个字符串中的一个字符。
示例 1:
输入: word1 = "sea", word2 = "eat" 输出: 2 解释: 第一步将 "sea" 变为 "ea" ,第二步将 "eat "变为 "ea"
示例 2:
输入:word1 = "leetcode", word2 = "etco" 输出:4
提示:
1 <= word1.length, word2.length <= 500
word1
和word2
只包含小写英文字母
定义 dp[i][j]
表示使得 word1[0:i-1]
和 word1[0:j-1]
两个字符串相同所需执行的删除操作次数。
时间复杂度:$O(mn)$。
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
m, n = len(word1), len(word2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
dp[i][0] = i
for j in range(1, n + 1):
dp[0][j] = j
for i in range(1, m + 1):
for j in range(1, n + 1):
if word1[i - 1] == word2[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1])
return dp[-1][-1]
class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length(), n = word2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; 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) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
}
}
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 <= 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) {
if (word1[i - 1] == word2[j - 1])
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[m][n];
}
};
func minDistance(word1 string, word2 string) int {
m, n := len(word1), len(word2)
dp := make([][]int, m+1)
for i := range dp {
dp[i] = make([]int, n+1)
dp[i][0] = i
}
for j := range dp[0] {
dp[0][j] = j
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
if word1[i-1] == word2[j-1] {
dp[i][j] = dp[i-1][j-1]
} else {
dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])
}
}
}
return dp[m][n]
}
function minDistance(word1: string, word2: string): number {
const m = word1.length;
const n = word2.length;
const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (word1[i - 1] === word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
const max = dp[m][n];
return m - max + n - max;
}
impl Solution {
pub fn min_distance(word1: String, word2: String) -> i32 {
let (m, n) = (word1.len(), word2.len());
let (word1, word2) = (word1.as_bytes(), word2.as_bytes());
let mut dp = vec![vec![0; n + 1]; m + 1];
for i in 1..=m {
for j in 1..=n {
dp[i][j] = if word1[i - 1] == word2[j - 1] {
dp[i - 1][j - 1] + 1
} else {
dp[i - 1][j].max(dp[i][j - 1])
};
}
}
let max = dp[m][n];
(m - max + (n - max)) as i32
}
}