Edit Distance
描述
Given two words word1
and word2
, find the minimum number of steps required to convert word1
to word2
. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
分析
设状态为f[i][j]
,表示A[0,i]
和B[0,j]
之间的最小编辑距离。设A[0,i]
的形式是str1c
,B[0,j]
的形式是str2d
,
- 如果
c==d
,则f[i][j]=f[i-1][j-1]
; 如果
c!=d
,- 如果将c替换成d,则
f[i][j]=f[i-1][j-1]+1
; - 如果在c后面添加一个d,则
f[i][j]=f[i][j-1]+1
; - 如果将c删除,则
f[i][j]=f[i-1][j]+1
;
- 如果将c替换成d,则
动规
// Edit Distance
// 二维动规,时间复杂度O(n*m),空间复杂度O(n*m)
class Solution {
public:
int minDistance(const string &word1, const string &word2) {
const size_t n = word1.size();
const size_t m = word2.size();
// 长度为n的字符串,有n+1个隔板
int f[n + 1][m + 1];
for (size_t i = 0; i <= n; i++)
f[i][0] = i;
for (size_t j = 0; j <= m; j++)
f[0][j] = j;
for (size_t i = 1; i <= n; i++) {
for (size_t j = 1; j <= m; j++) {
if (word1[i - 1] == word2[j - 1])
f[i][j] = f[i - 1][j - 1];
else {
int mn = min(f[i - 1][j], f[i][j - 1]);
f[i][j] = 1 + min(f[i - 1][j - 1], mn);
}
}
}
return f[n][m];
}
};
动规+滚动数组
// Edit Distance
// 二维动规+滚动数组
// 时间复杂度O(n*m),空间复杂度O(n)
class Solution {
public:
int minDistance(const string &word1, const string &word2) {
if (word1.length() < word2.length())
return minDistance(word2, word1);
int f[word2.length() + 1];
int upper_left = 0; // 额外用一个变量记录f[i-1][j-1]
for (size_t i = 0; i <= word2.size(); ++i)
f[i] = i;
for (size_t i = 1; i <= word1.size(); ++i) {
upper_left = f[0];
f[0] = i;
for (size_t j = 1; j <= word2.size(); ++j) {
int upper = f[j];
if (word1[i - 1] == word2[j - 1])
f[j] = upper_left;
else
f[j] = 1 + min(upper_left, min(f[j], f[j - 1]));
upper_left = upper;
}
}
return f[word2.length()];
}
};