From 82468b6f2ae5e399306e3deedb0af540671b5417 Mon Sep 17 00:00:00 2001
From: markwang <www.wang61@qq.com>
Date: Tue, 22 Oct 2024 11:31:15 +0800
Subject: [PATCH] =?UTF-8?q?583.=E4=B8=A4=E4=B8=AA=E5=AD=97=E7=AC=A6?=
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---
 .../0583.两个字符串的删除操作.md    | 33 ++++++++++++++++++-
 1 file changed, 32 insertions(+), 1 deletion(-)

diff --git a/problems/0583.两个字符串的删除操作.md b/problems/0583.两个字符串的删除操作.md
index 14a55631..57bd3abe 100644
--- a/problems/0583.两个字符串的删除操作.md
+++ b/problems/0583.两个字符串的删除操作.md
@@ -33,7 +33,7 @@
 
 dp[i][j]：以i-1为结尾的字符串word1，和以j-1位结尾的字符串word2，想要达到相等，所需要删除元素的最少次数。 
 
-这里dp数组的定义有点点绕，大家要撸清思路。
+这里dp数组的定义有点点绕，大家要理清思路。
 
 2. 确定递推公式
 
@@ -255,6 +255,8 @@ class Solution(object):
 ```
 ### Go：
 
+动态规划一
+
 ```go
 func minDistance(word1 string, word2 string) int {
 	dp := make([][]int, len(word1)+1)
@@ -287,6 +289,35 @@ func min(a, b int) int {
 	return b
 }
 ```
+
+动态规划二
+
+```go
+func minDistance(word1 string, word2 string) int {
+    dp := make([][]int, len(word1) + 1)
+    for i := range dp {
+        dp[i] = make([]int, len(word2) + 1)
+    }
+    for i := 1; i <= len(word1); i++ {
+        for j := 1; j <= len(word2); j++ {
+            if word1[i-1] == word2[j-1] {
+                dp[i][j] = dp[i-1][j-1] + 1
+            } else {
+                dp[i][j] = max(dp[i-1][j], dp[i][j-1])
+            }
+        }
+    }
+    return len(word1) + len(word2) - dp[len(word1)][len(word2)] * 2
+}
+
+func max(x, y int) int {
+    if x > y {
+        return x
+    }
+    return y
+}
+```
+
 ### Javascript：
 
 ```javascript