From 4d6b8252a1d51b0cef64e5b4739ef21329bdd41c Mon Sep 17 00:00:00 2001
From: chenzhg <2216468566@qq.com>
Date: Thu, 29 Sep 2022 17:31:58 +0800
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---
 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);
                 }
             }
         }