From 8f269ef2b1c2da6551dcf97f8c21cf50387bf447 Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Tue, 5 Mar 2024 11:09:20 +0800
Subject: [PATCH] add leetcode content

---
 content/posts/leetcode.md | 51 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 51 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index d5fce6f..6f359f1 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -518,3 +518,54 @@ func bagOfTokensScore(tokens []int, power int) int {
 ### 总结
 
 在实现过程中, 起初使用一个数组来保存每次的score值, 这样空间复杂度略大, 后来查看他人代码, 发现只需要一个max变量来保存当前最大的score值, 并在每次循环计算当前轮次的score值时与当前的最大值比较并根据二者大小更新max变量的值即可, 这样只需要O(1)的空间复杂度.
+
+## day8 2024-03-05
+
+### 1750. Minimum Length of String After Deleting Similar Ends
+
+Given a string s consisting only of characters 'a', 'b', and 'c'. You are asked to apply the following algorithm on the string any number of times:
+
+1. Pick a non-empty prefix from the string s where all the characters in the prefix are equal.
+2. Pick a non-empty suffix from the string s where all the characters in this suffix are equal.
+3. The prefix and the suffix should not intersect at any index.
+4. The characters from the prefix and suffix must be the same.
+5. Delete both the prefix and the suffix.
+   Return the minimum length of s after performing the above operation any number of times (possibly zero times).
+
+![0305Qc2qNiDBOKRk](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0305Qc2qNiDBOKRk.png)
+
+### 题解
+
+本题的题目略显复杂, 读起来乍一看让人不明所以, 实际上题目的目标即为从字符串的前后同时删除相同的字符, 删除时只要是相同的就全部删除, 如最前面有一个a最后面有3个a则同时将这四个a删除. 删除的字符串下标不能相交. 思路比较简单, 判断字符串的前后字符是否相同, 然后删除前后的连续相同字符即可, 注意下标不要重叠. 同时注意边界情况, 字符串只有一个字符的情况单独处理一下(直接返回1).
+
+### 代码
+
+```go
+func minimumLength(s string) int {
+    forward := 0
+    backward := 0
+    for s[0] == s[len(s)-1]{
+    if len(s) == 1{
+        return 1
+    } else {
+            forward = 0
+            backward = len(s)-1
+            for forward<backward && s[forward+1] == s[forward] {
+                forward++
+            }
+            for backward>forward && s[backward-1] == s[backward] {
+                backward--
+            }
+            if forward == backward{
+                return 0
+            }
+            s = s[forward+1:backward]
+        }
+    }
+    return len(s)
+}
+```
+
+### 总结
+
+本题值得注意的地方在于for循环中条件的设置, 可能会忽略与运算符两侧条件表达式的前后顺序, 但由于短路机制的存在, 与运算符两侧表达式的前后顺序有时非常重要, 例如在本题中, 如果将forward<backward条件设置在前面, 则当forward到达数组的边界时会直接退出循环, 但若将相等判断放在前面, 会因为当forward到达数组边界时forward+1下标越界而出错.