From 4d6b8252a1d51b0cef64e5b4739ef21329bdd41c Mon Sep 17 00:00:00 2001
From: chenzhg <2216468566@qq.com>
Date: Thu, 29 Sep 2022 17:31:58 +0800
Subject: [PATCH 1/2] =?UTF-8?q?Update=200583.=E4=B8=A4=E4=B8=AA=E5=AD=97?=
 =?UTF-8?q?=E7=AC=A6=E4=B8=B2=E7=9A=84=E5=88=A0=E9=99=A4=E6=93=8D=E4=BD=9C?=
 =?UTF-8?q?=20C++?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0583.两个字符串的删除操作.md | 4 +++-
 1 file changed, 3 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 fd80853e..c33f7f58 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
@@ -47,6 +47,8 @@ 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word
 
 那最后当然是取最小值，所以当word1[i - 1] 与 word2[j - 1]不相同的时候，递推公式：dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
 
+因为dp[i - 1][j - 1] + 1等于 dp[i - 1][j] 或 dp[i][j - 1]，所以递推公式可简化为：dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
+
 
 3. dp数组如何初始化
 
@@ -90,7 +92,7 @@ public:
                 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] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
+                    dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
                 }
             }
         }

From e4ca3482fbb5202172c6268a10565dfbf116f6fc Mon Sep 17 00:00:00 2001
From: chenzhg <2216468566@qq.com>
Date: Sat, 1 Oct 2022 16:25:03 +0800
Subject: [PATCH 2/2] =?UTF-8?q?Update=200072.=E7=BC=96=E8=BE=91=E8=B7=9D?=
 =?UTF-8?q?=E7=A6=BB=20=E6=B7=BB=E5=8A=A0C=E8=AF=AD=E8=A8=80=E7=89=88?=
 =?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0072.编辑距离.md | 27 +++++++++++++++++++++++++++
 1 file changed, 27 insertions(+)

diff --git a/problems/0072.编辑距离.md b/problems/0072.编辑距离.md
index e40461de..a15012d7 100644
--- a/problems/0072.编辑距离.md
+++ b/problems/0072.编辑距离.md
@@ -362,6 +362,33 @@ function minDistance(word1: string, word2: string): number {
 };
 ```
 
+C：
+
+```c
+int min(int num1, int num2, int num3) {
+    return num1 > num2 ? (num2 > num3 ? num3 : num2) : (num1 > num3 ? num3 : num1); 
+}
+
+int minDistance(char * word1, char * word2){
+    int dp[strlen(word1)+1][strlen(word2)+1];
+    dp[0][0] = 0;
+    for (int i = 1; i <= strlen(word1); i++) dp[i][0] = i;
+    for (int i = 1; i <= strlen(word2); i++) dp[0][i] = i;
+
+    for (int i = 1; i <= strlen(word1); i++) {
+        for (int j = 1; j <=strlen(word2); j++) {
+            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][j-1], dp[i-1][j]) + 1;
+            }
+        }
+    }
+    return dp[strlen(word1)][strlen(word2)];
+}
+```
+
 
 
 -----------------------