def min_operations(x, y): m, n = len(x), len(y) dp = [[0] * (n+1) for _ in range(m+1)] for i in range(m+1): dp[i][0] = i for j in range(n+1): dp[0][j] = j for i in range(1, m+1): for j in range(1, n+1): if x[i-1] == y[j-1]: dp[i][j] = dp[i-1][j-1] else: dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1 return dp[m][n]转换为c++语言
时间: 2024-04-24 14:21:41 浏览: 15
以下是将`min_operations`函数转换为C++语言的代码:
```cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int min_operations(string x, string y) {
int m = x.length(), n = y.length();
vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
for (int i = 0; i <= m; i++) {
dp[i][0] = i;
}
for (int j = 0; j <= n; j++) {
dp[0][j] = j;
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (x[i-1] == y[j-1]) {
dp[i][j] = dp[i-1][j-1];
} else {
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1;
}
}
}
return dp[m][n];
}
int main() {
string x = "101";
string y = "11010";
int result = min_operations(x, y);
cout << "Minimum operations required: " << result << endl;
return 0;
}
```
你可以将`x`和`y`替换为你想要的输入,然后运行这段C++代码来获取所需的最少操作次数。
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