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Jan 18, 2025
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[pull] main from doocs:main #311

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142 changes: 86 additions & 56 deletions solution/0500-0599/0583.Delete Operation for Two Strings/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -56,11 +56,15 @@ tags:

### 方法一：动态规划

类似[1143. 最长公共子序列](https://github.com/doocs/leetcode/blob/main/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md)。
我们定义 $f[i][j]$ 表示使得字符串 $\textit{word1}$ 的前 $i$ 个字符和字符串 $\textit{word2}$ 的前 $j$ 个字符相同的最小删除步数。那么答案为 $f[m][n]$，其中 $m$ 和 $n$ 分别是字符串 $\textit{word1}$ 和 $\textit{word2}$ 的长度。

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。
初始时，如果 $j = 0$，那么 $f[i][0] = i$；如果 $i = 0$，那么 $f[0][j] = j$。

时间复杂度：$O(mn)$。
当 $i, j > 0$ 时，如果 $\textit{word1}[i - 1] = \textit{word2}[j - 1]$，那么 $f[i][j] = f[i - 1][j - 1]$；否则 $f[i][j] = \min(f[i - 1][j], f[i][j - 1]) + 1$。

最终返回 $f[m][n]$ 即可。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是字符串 $\textit{word1}$ 和 $\textit{word2}$ 的长度。

<!-- tabs:start -->

Expand All @@ -70,18 +74,18 @@ tags:
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)]
f = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
dp[i][0] = i
f[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]
f[0][j] = j
for i, a in enumerate(word1, 1):
for j, b in enumerate(word2, 1):
if a == b:
f[i][j] = f[i - 1][j - 1]
else:
dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1])
return dp[-1][-1]
f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1
return f[m][n]
```

#### Java
Expand All @@ -90,23 +94,25 @@ class Solution:
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;
int[][] f = new int[m + 1][n + 1];
for (int i = 0; i <= m; ++i) {
f[i][0] = i;
}
for (int j = 1; j <= n; ++j) {
dp[0][j] = j;
for (int j = 0; j <= n; ++j) {
f[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];
char a = word1.charAt(i - 1);
char b = word2.charAt(j - 1);
if (a == b) {
f[i][j] = f[i - 1][j - 1];
} else {
dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + 1;
}
}
}
return dp[m][n];
return f[m][n];
}
}
```
Expand All @@ -117,19 +123,26 @@ class Solution {
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;
int m = word1.length(), n = word2.length();
vector<vector<int>> f(m + 1, vector<int>(n + 1));
for (int i = 0; i <= m; ++i) {
f[i][0] = i;
}
for (int j = 0; j <= n; ++j) {
f[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]);
char a = word1[i - 1];
char b = word2[j - 1];
if (a == b) {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1;
}
}
}
return dp[m][n];
return f[m][n];
}
};
```
Expand All @@ -139,24 +152,25 @@ public:
```go
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
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
f[i][0] = i
}
for j := range dp[0] {
dp[0][j] = j
for j := 1; j <= n; j++ {
f[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]
a, b := word1[i-1], word2[j-1]
if a == b {
f[i][j] = f[i-1][j-1]
} else {
dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])
f[i][j] = 1 + min(f[i-1][j], f[i][j-1])
}
}
}
return dp[m][n]
return f[m][n]
}
```

Expand All @@ -166,18 +180,23 @@ func minDistance(word1 string, word2 string) int {
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++) {
const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
for (let i = 1; i <= m; ++i) {
f[i][0] = i;
}
for (let j = 1; j <= n; ++j) {
f[0][j] = j;
}
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;
f[i][j] = f[i - 1][j - 1];
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + 1;
}
}
}
const max = dp[m][n];
return m - max + n - max;
return f[m][n];
}
```

Expand All @@ -186,20 +205,31 @@ function minDistance(word1: string, word2: string): number {
```rust
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];
let m = word1.len();
let n = word2.len();
let s: Vec<char> = word1.chars().collect();
let t: Vec<char> = word2.chars().collect();
let mut f = vec![vec![0; n + 1]; m + 1];

for i in 0..=m {
f[i][0] = i as i32;
}
for j in 0..=n {
f[0][j] = j as i32;
}

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
let a = s[i - 1];
let b = t[j - 1];
if a == b {
f[i][j] = f[i - 1][j - 1];
} else {
dp[i - 1][j].max(dp[i][j - 1])
};
f[i][j] = std::cmp::min(f[i - 1][j], f[i][j - 1]) + 1;
}
}
}
let max = dp[m][n];
(m - max + (n - max)) as i32
f[m][n]
}
}
```
Expand Down
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