https://leetcode-cn.com/problems/edit-distance/description/
思路如下：
计算编辑距离
| : | : | : | : | : | : |
|---|---|---|---|---|---|
| - | - | a | d | i | t |
| - | 0 | 1 | 2 | 3 | 4 |
| e | 1 | 1 | |||
| d | 2 | ||||
| a | 3 | ||||
| t | 4 | ||||
| x | 5 |
old（记录左上角的值)
动态规划的思路很简单， 倒推
python代码
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
len1, len2 = len(word1), len(word2)
if len1 == 0 or len2 == 0:
return max(len1, len2)
dp = [[0] * (len2 + 1) for _ in range(len1 + 1)] # len1行 len2列
for i in range(len1 + 1):
dp[i][0] = i;
for j in range(len2 + 1):
dp[0][j] = j
for i in range(1, len1 + 1):
for j in range(1, len2 + 1):
if(word1[i-1] == word2[j-1]):
dp[i][j] = dp[i-1][j-1];
else:
dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1
return dp[len1][len2]
scala代码
// 计算编辑距离
/*
a d i t old（记录左上角的值, pw(0) 更新为)
0 1 2 3 4
e 1 1 0
d 2
a 3
t 4
x
*/
object Solution {
def minDistance(x: String, y: String): Int = {
val pw = new Array[Int](y.length + 1)
for(j <- Range(0, y.length + 1)) pw(j) = j
for(i <- Range(1, x.length + 1))
{
var old = i - 1 // 初始左上角的值是i - 1
pw(0) = i // 初始最左边的值是 i, 最上面的值是 j x初始最左边的值
for(j <- Range(1, y.length + 1)){
val temp = pw(j)
if(x(i - 1) == y(j - 1)){
pw(j) = old
}
else{
pw(j) = math.min(math.min(pw(j) + 1, pw(j-1) + 1), old + 1)
}
old = temp // pw(j)更新后, old替换为pw(j)原来的值,也就是左上角的值
}
}
pw(y.length)
}
}
