Refine 541_Reverse_String_II

Author: qiyuangong

README.md

@@ -164,6 +164,7 @@ I'm currently working on [Analytics-Zoo](https://github.com/intel-analytics/anal
 | 485 | [Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/485_Max_Consecutive_Ones.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/485_Max_Consecutive_Ones.java) | Add one when encounter 1, set to 0 when encounter 0, O(n) and O(1) |
 | 509 | [Fibonacci Number](https://leetcode.com/problems/fibonacci-number/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/509_Fibonacci_Number.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/509_Fibonacci_Number.java) | 1. Recursive, O(n) <br>2. DP with memo, O(n). Note that N<=30, which means that we can keep a memo from 0 to 30. |
 | 538 | [Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/538_Convert_BST_to_Greater_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/538_Convert_BST_to_Greater_Tree.java) | Right first DFS with a variable recording sum of node.val and right.val. 1. Recursive.<br>2. Stack 3. Reverse Morris In-order Traversal |
+| 541 | [Reverse String II](https://leetcode.com/problems/reverse-string-ii/solution/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/541_Reverse_String_II.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/541_Reverse_String_II.java) | Handle each 2k until reaching end, On(n) and O(n) |
 | 543 | [Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/543_Diameter_of_Binary_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/543_Diameter_of_Binary_Tree.java) | DFS with O(1) for max answer |
 | 547 | [Friend Circles](https://leetcode.com/problems/friend-circles/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/547_Friend_Circles.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/547_Friend_Circles.java) | 1. DFS, O(n^2) and O(1)<br>2. BFS, O(n^2) and O(1)<br>3. Union-find, O(n^2) and O(n)|
 | 557 | [Reverse Words in a String III](https://leetcode.com/problems/reverse-words-in-a-string-iii/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/557_Reverse_Words_in_a_String_III.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/557_Reverse_Words_in_a_String_III.java) | String handle: Split with space than reverse word, O(n) and O(n). Better solution is that reverse can be O(1) space in array. |

java/541_Reverse_String_II.java

@@ -0,0 +1,16 @@
+class Solution {
+    public String reverseStr(String s, int k) {
+        // https://leetcode.com/problems/reverse-string-ii/solution/
+        char[] a = s.toCharArray();
+        for (int start = 0; start < a.length; start += 2 * k) {
+            int i = start, j = Math.min(start + k - 1, a.length - 1);
+            // Reverse from i to j
+            while (i < j) {
+                char tmp = a[i];
+                a[i++] = a[j];
+                a[j--] = tmp;
+            }
+        }
+        return new String(a);
+    }
+}

python/541_Reverse_String_II.py

@@ -1,5 +1,5 @@
 class Solution:
-    def reverseStr(self, s, k):
+    def reverseStr(self, s: str, k: int) -> str:
         N = len(s)
         ans = ""
         position = 0
@@ -9,6 +9,12 @@ def reverseStr(self, s, k):
             position += 2 * k
         return ans
 
+    # def reverseStr(self, s: str, k: int) -> str:
+    #     s = list(s)
+    #     for i in range(0, len(s), 2*k):
+    #         s[i:i+k] = reversed(s[i:i+k])
+    #     return "".join(s)
+
 
 
 s1 = Solution()