【Leetcode】72.編輯距離
阿新 • • 發佈:2018-12-11
題目
給定兩個單詞 word1 和 word2,計算出將 word1 轉換成 word2 所使用的最少運算元 。你可以對一個單詞進行如下三種操作:
插入一個字元
刪除一個字元
替換一個字元
示例 1:
輸入: word1 = "horse", word2 = "ros"
輸出: 3
解釋:
horse -> rorse (將 'h' 替換為 'r')
rorse -> rose (刪除 'r')
rose -> ros (刪除 'e')
示例 2:
輸入: word1 = "intention", word2 = "execution" 輸出: 5 解釋: intention -> inention (刪除 't') inention -> enention (將 'i' 替換為 'e') enention -> exention (將 'n' 替換為 'x') exention -> exection (將 'n' 替換為 'c') exection -> execution (插入 'u')
題解
這個題目拿到題目基本就能想到DP,因為感覺和我們之前的爬樓梯啥的比較相似。這個題目比較為hard主要是,狀態轉換比較複雜。
定義: dpi , word1這個字串的前i個 -> word1這個字串的前j 個字元,所需要的最小的步數
那麼有以下幾種情況
word1[i] == word2[j]
不需要做變化,那麼 dpi = dp[i-1,j-1]
word1[i] != word2[j]
我們就需要動用上面那三種操作了:
- add = dp[i, j-1], 代表插入一個字元
- delete = dp[i-1, j],代表刪除一個字元
- replace = dp[i-1, j-1],代表替換一個字元
- dpi = 1 + min(add, delete, replace)
時間複雜度 o(m * n)
java
class Solution { public int minDistance(String word1, String word2) { int m = word1.length(); int n = word2.length(); int[][] dp = new int[m + 1][n + 1]; // base 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 = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (word1.charAt(i) == word2.charAt(j)) { dp[i + 1][j + 1] = dp[i][j]; } else { int add = dp[i][j + 1]; int delete = dp[i + 1][j]; int rep = dp[i][j]; dp[i + 1][j + 1] = Math.min(Math.min(add, delete), rep) + 1; } } } return dp[m][n]; } }
python
class Solution:
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
m = len(word1)
n = len(word2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(m + 1):
dp[i][0] = i
for i in range(n + 1):
dp[0][i] = i
for i in range(m):
for j in range(n):
if word1[i] == word2[j]:
dp[i + 1][j + 1] = dp[i][j]
else:
add = dp[i][j + 1]
delete = dp[i + 1][j]
replace = dp[i][j]
dp[i + 1][j + 1] = min(add, delete, replace) + 1
return dp[m][n]
近期文章