From d85bfc486cc5270e17f4efa886e5a8c3fc936435 Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Tue, 15 Oct 2024 12:34:23 +0800
Subject: [PATCH] leetcode update

---
 content/posts/leetcode.md | 41 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 41 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 463e931..6cad86d 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -15634,3 +15634,44 @@ public:
     }
 };
 ```
+## day231 2024-10-15
+### 2938. Separate Black and White Balls 
+There are n balls on a table, each ball has a color black or white.
+
+You are given a 0-indexed binary string s of length n, where 1 and 0 represent black and white balls, respectively.
+
+In each step, you can choose two adjacent balls and swap them.
+
+Return the minimum number of steps to group all the black balls to the right and all the white balls to the left. 
+
+![101581cgwXkBIJs6](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/101581cgwXkBIJs6.png)
+
+### 题解
+本题起初想到可以遍历一遍数组统计0的个数，再遍历数组将前面的1和后面的0交换直到到达0的个数的位置。但这样好像麻烦了一些，统计0的个数其实没有必要，直接使用双指针，分别从首尾开始遍历数组，尾指针遇到的0和首指针遇到的1交换，直到两个指针相遇即可。原因在于尾指针每次遇到0就将其和1交换，则尾指针指向位置之后的数组可以确保是全1的，同理首指针之前的数组可以确保是全0的。二者相遇时相遇位置之后的数组为全1，之前的为全0，就已经满足题目条件了。
+
+遍历过程中，尾指针遇到0停下，移动首指针直到遇到1，计算二者距离并加和到结果中，继续移动尾指针直到头尾相遇结束。
+
+### 代码
+```cpp 
+class Solution {
+public:
+    long long minimumSteps(string s) {
+        long long int head = 0;
+        long long int tail = s.size()-1;
+        long long int result = 0;
+        while (tail > head){
+            while(s[tail] != '0' && tail > head){
+                tail--;
+            }
+            while(s[head] != '1' && head < tail){
+                head++;
+            }
+            result += tail-head;
+            tail--;
+            head++;
+        }
+        return result;
+    }
+};
+```
+