1. 题目来源
链接:72. 编辑距离
2. 题目解析
前导题:
最经典的 dp
题之一了吧。上面的前导题已经讲解的很清楚了!
- 时间复杂度: O ( n 2 ) O(n^2) O(n2)。
- 空间复杂度: O ( n 2 ) O(n^2) O(n2)
代码:
class Solution {
public:
int minDistance(string word1, string word2) {
int n = word1.size(), m = word2.size();
word1 = ' ' + word1, word2 = ' ' + word2;
vector<vector<int>> f(n + 1, vector<int>(m + 1, 1e8));
// 若A串或B串为空串的时候,那么就将字符全部添加即可,即当前字符串的长度
for (int i = 0; i <= n; i ++ ) f[i][0] = i;
for (int i = 0; i <= m; i ++ ) f[0][i] = i;
for (int i = 1; i <= n; i ++ ) {
for (int j = 1; j <= m; j ++ ) {
f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1;
int t = word1[i] != word2[j];
f[i][j] = min(f[i][j], f[i - 1][j - 1] + t);
}
}
return f[n][m];
}
};