feat: update solutions to lc problem: No.1888. Minimum Number of Flips to Make the Binary String Alternating

Author: yanglbme

solution/0300-0399/0328.Odd Even Linked List/Solution.go

@@ -5,18 +5,18 @@
  *     Next *ListNode
  * }
  */
- func oddEvenList(head *ListNode) *ListNode {
-    if head == nil {
-        return head
-    }
-    odd, even := head, head.Next
-    evenHead := even
-    for even != nil && even.Next != nil {
-        odd.Next = even.Next
-        odd = odd.Next
-        even.Next = odd.Next
-        even = even.Next
-    }
-    odd.Next = evenHead
-    return head
+func oddEvenList(head *ListNode) *ListNode {
+	if head == nil {
+		return head
+	}
+	odd, even := head, head.Next
+	evenHead := even
+	for even != nil && even.Next != nil {
+		odd.Next = even.Next
+		odd = odd.Next
+		even.Next = odd.Next
+		even = even.Next
+	}
+	odd.Next = evenHead
+	return head
 }

solution/1800-1899/1888.Minimum Number of Flips to Make the Binary String Alternating/README.md

@@ -58,7 +58,7 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-“双倍字符串 + 滑动窗口”实现。
+“滑动窗口”实现。
 
 <!-- tabs:start -->
 
@@ -75,10 +75,9 @@ class Solution:
         for i, c in enumerate(s):
             cnt += c != target[i & 1]
         res = min(cnt, n - cnt)
-        s *= 2
         for i in range(n):
             cnt -= s[i] != target[i & 1]
-            cnt += s[i + n] != target[(i + n) & 1]
+            cnt += s[i] != target[(i + n) & 1]
             res = min(res, cnt, n - cnt)
         return res
 ```
@@ -97,10 +96,9 @@ class Solution {
             cnt += (s.charAt(i) == target.charAt(i & 1) ? 0 : 1);
         }
         int res = Math.min(cnt, n - cnt);
-        s += s;
         for (int i = 0; i < n; ++i) {
             cnt -= (s.charAt(i) == target.charAt(i & 1) ? 0 : 1);
-            cnt += (s.charAt(i + n) == target.charAt((i + n) & 1) ? 0 : 1);
+            cnt += (s.charAt(i) == target.charAt((i + n) & 1) ? 0 : 1);
             res = Math.min(res, Math.min(cnt, n - cnt));
         }
         return res;

solution/1800-1899/1888.Minimum Number of Flips to Make the Binary String Alternating/README_EN.md

@@ -69,10 +69,9 @@ class Solution:
         for i, c in enumerate(s):
             cnt += c != target[i & 1]
         res = min(cnt, n - cnt)
-        s *= 2
         for i in range(n):
             cnt -= s[i] != target[i & 1]
-            cnt += s[i + n] != target[(i + n) & 1]
+            cnt += s[i] != target[(i + n) & 1]
             res = min(res, cnt, n - cnt)
         return res
 ```
@@ -89,10 +88,9 @@ class Solution {
             cnt += (s.charAt(i) == target.charAt(i & 1) ? 0 : 1);
         }
         int res = Math.min(cnt, n - cnt);
-        s += s;
         for (int i = 0; i < n; ++i) {
             cnt -= (s.charAt(i) == target.charAt(i & 1) ? 0 : 1);
-            cnt += (s.charAt(i + n) == target.charAt((i + n) & 1) ? 0 : 1);
+            cnt += (s.charAt(i) == target.charAt((i + n) & 1) ? 0 : 1);
             res = Math.min(res, Math.min(cnt, n - cnt));
         }
         return res;

solution/1800-1899/1888.Minimum Number of Flips to Make the Binary String Alternating/Solution.java

@@ -7,10 +7,9 @@ public int minFlips(String s) {
             cnt += (s.charAt(i) == target.charAt(i & 1) ? 0 : 1);
         }
         int res = Math.min(cnt, n - cnt);
-        s += s;
         for (int i = 0; i < n; ++i) {
             cnt -= (s.charAt(i) == target.charAt(i & 1) ? 0 : 1);
-            cnt += (s.charAt(i + n) == target.charAt((i + n) & 1) ? 0 : 1);
+            cnt += (s.charAt(i) == target.charAt((i + n) & 1) ? 0 : 1);
             res = Math.min(res, Math.min(cnt, n - cnt));
         }
         return res;

solution/1800-1899/1888.Minimum Number of Flips to Make the Binary String Alternating/Solution.py

@@ -6,9 +6,8 @@ def minFlips(self, s: str) -> int:
         for i, c in enumerate(s):
             cnt += c != target[i & 1]
         res = min(cnt, n - cnt)
-        s *= 2
         for i in range(n):
             cnt -= s[i] != target[i & 1]
-            cnt += s[i + n] != target[(i + n) & 1]
+            cnt += s[i] != target[(i + n) & 1]
             res = min(res, cnt, n - cnt)
         return res