Update 072._edit_distance.md

docs/Leetcode_Solutions/072._edit_distance.md

-### 72. Edit Distance
+# 72. Edit Distance
 
-题目:
-<https://leetcode.com/problems/edit-distance/>
+**<font color=red>难度: Hard</font>**
 
+## 刷题内容
 
-难度:
+> 原题连接
 
-Hard
+* https://leetcode.com/problems/edit-distance/description/
+
+> 内容描述
+
+```
+Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
+
+You have the following 3 operations permitted on a word:
+
+Insert a character
+Delete a character
+Replace a character
+Example 1:
+
+Input: word1 = "horse", word2 = "ros"
+Output: 3
+Explanation: 
+horse -> rorse (replace 'h' with 'r')
+rorse -> rose (remove 'r')
+rose -> ros (remove 'e')
+Example 2:
+
+Input: word1 = "intention", word2 = "execution"
+Output: 5
+Explanation: 
+intention -> inention (remove 't')
+inention -> enention (replace 'i' with 'e')
+enention -> exention (replace 'n' with 'x')
+exention -> exection (replace 'n' with 'c')
+exection -> execution (insert 'u')
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(len(word1)*len(word2))******- 空间复杂度: O(len(word1)*len(word2))******
 
 可以做的操作：
 
@@ -48,39 +83,7 @@ Hard
 
 ***要始终明确一点，```dp[i][j]```的含义是使得```word1的前i字符子串```与```word2的前j字符子串```相等所需要的操作数，这也是为什么我们需要在初始化```dp矩阵```时需要行列数均加上```1```***
 
-用wikipedia上的伪码改造
 
-```
-function LevenshteinDistance(char s[1..m], char t[1..n]):
-  // for all i and j, d[i,j] will hold the Levenshtein distance between
-  // the first i characters of s and the first j characters of t
-  // note that d has (m+1)*(n+1) values
-  declare int d[0..m, 0..n]
- 
-  set each element in d to zero
- 
-  // source prefixes can be transformed into empty string by
-  // dropping all characters
-  for i from 1 to m:
-      d[i, 0] := i
- 
-  // target prefixes can be reached from empty source prefix
-  // by inserting every character
-  for j from 1 to n:
-      d[0, j] := j
- 
-  for j from 1 to n:
-      for i from 1 to m:
-          if s[i] = t[j]:
-            substitutionCost := 0
-          else:
-            substitutionCost := 1
-          d[i, j] := minimum(d[i-1, j] + 1,                   // deletion
-                             d[i, j-1] + 1,                   // insertion
-                             d[i-1, j-1] + substitutionCost)  // substitution
- 
-  return d[m, n]
-```
 
 对应的例子表格图