feat(invidam): week 2 update.

invert-binary-tree/invidam.go.md

@@ -0,0 +1,91 @@
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+Use DFS, referencing the property of a tree (i.e. child node is the root node of a subtree.)
+# Approach
+<!-- Describe your approach to solving the problem. -->
+1. Visit the root node.
+2. If visited node is `nil`, return `nil`
+3. Swap left and right nodes.
+4. Visit Swapped left and right nodes.
+5. Repeat Step 2 ~ 4.
+# Complexity
+- Time complexity: $$O(n)$$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(n)$$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+
+# Code
+```
+func invertTree(root *TreeNode) *TreeNode {
+    if root == nil {
+        return nil
+    }
+    root.Left, root.Right = invertTree(root.Right), invertTree(root.Left)
+    return root
+}
+```
+- - -
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+Visit, But Use BFS. (`for loop`)
+# Approach
+<!-- Describe your approach to solving the problem. -->
+1. Create Queue and push root node to it.
+2. If the queue is empty, return the `root` node.
+3. Otherwise, pop the top node.
+4. Swap the left and right children of the removed node.
+5. Push swapped children.
+4. Repeat Step
+# Complexity
+- Time complexity: $$O(n)$$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(n)$$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+
+# Code
+```
+type Queue[T any] struct {
+    Index int
+    Nodes []T
+}
+
+func NewQueue[T any]() Queue[T] {
+    return Queue[T]{Nodes: make([]T, 0)}
+}
+
+func (q *Queue[T]) Push(node T) {
+    q.Nodes = append(q.Nodes, node)    
+}
+
+func (q *Queue[T]) Pop() T {
+    ret := q.Nodes[q.Index]
+    q.Index++
+    return ret
+}
+
+func (q *Queue[T]) IsEmpty() bool {
+    return q.Index == len(q.Nodes)
+}
+
+func invertTree(root *TreeNode) *TreeNode {
+    q := NewQueue[*TreeNode]()
+    q.Push(root)
+    for !q.IsEmpty() {
+        t := q.Pop()
+        if t == nil {
+            continue
+        }
+        t.Left, t.Right = t.Right, t.Left
+        
+        q.Push(t.Left)
+        q.Push(t.Right)
+    }
+    return root
+}
+```
+
+# Learned
+- 고언어에서 `a, b = b, a` 처럼 간결한 코딩이 가능하다.
+- 포인터 타입과 일반 타입의 차이 (고언어에서는 일반 타입을 넘길 시 무조건 복사한다.)

linked-list-cycle/invidam.go.md

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+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+Makr elements as visited by setting a dirty bit.
+Check Dirty bit to visited nodes.
+# Approach
+<!-- Describe your approach to solving the problem. -->
+1. Initiate a constant value named `visited` over the input range. (In my case `-10001`.)
+2. Start Visitng node. by head.
+3. Check if the current node is `nil` or its value matches the visited value (Initially, the head node should not be visited value)
+4. If node is valid(i.e. not `nil` and not `visited`), set the node's value as the visited value. 
+5. Move to the next node and repeat step 3.
+
+# Complexity
+- Time complexity: $$O(n)$$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+
+
+- Space complexity: $$O(n)$$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+(n = list's size)
+# Code
+```
+const visited = -10001
+func hasCycle(head *ListNode) bool {
+    if head == nil {
+        return false;
+    }
+    if head.Val == visited {
+        return true;
+    }
+    head.Val = visited
+    return hasCycle(head.Next)
+}
+```
+
+- - -
+# Institution
+Use "The tortoise and hare" Algorithom.
+# Approach
+<!-- Describe your approach to solving the problem. -->
+1. Designate two node to move at differnt speeds. One node should move faster (referred to as `fast`), and the other should move slower (`slow`)
+2. Allow both nodes to iterate through the graph, with each node moving at its designated speed.
+3. If the `fast` node catches up to the `slow` node (i.e. both node points to the same node at same point), the the graph contains a cycle.is cycle.
+# Complexity
+- Time complexity: $$O(n)$$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+- Space complexity: $$O(1)$$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+# Code
+```
+func hasCycle(head *ListNode) bool {
+    if head == nil || head.Next == nil {
+        return false
+    }
+    for slow, fast := head, head.Next; fast != nil && fast.Next != nil; slow, fast = slow.Next, fast.Next.Next{
+        if slow == fast {
+            return true
+        }
+    }
+    return false
+}
+
+```
+# Learned
+- 약한 부분이었던 투포인터 알고리즘에 대해 좀 더 생각해보았다.
+- 반복문을 깔끔하게 작성하는 법을 고민해보았다.

merge-two-sorted-lists/invidam.go.md

@@ -0,0 +1,116 @@
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+링크드리스트 --> 반복문 or 재귀호출이 떠올랐다.
+# Approach
+<!-- Describe your approach to solving the problem. -->
+- 재귀 호출과 반복문 사이에서 고민하였다.
+- 새로운 노드를 만들고, 작은 노드들을 이어 붙였다. (V1_1, V2_1)
+- 새로운 노드의 생성 없이, 기존 노드들을 연결하여 해결할 수도 있었다. (V1_2, V2_2)
+# Complexity
+- Time complexity: $$O(N+M)$$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(N+M)$$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+
+(N,M은 각각 두 링크드리스트의 길이)
+# Code
+```
+func mergeTwoListsV1(list1 *ListNode, list2 *ListNode) *ListNode {
+	if list1 == nil && list2 == nil {
+		return nil
+	} else if list1 != nil {
+		return list1
+	} else if list2 != nil {
+		return list2
+	}
+	// assert list1, list2 is not nil
+	var val int
+	if list1.Val < list2.Val {
+		val = list1.Val
+		list1 = list1.Next
+	} else {
+		val = list2.Val
+		list2 = list2.Next
+	}
+
+	return &ListNode{Val: val, Next: mergeTwoListsV1(list1, list2)}
+}
+
+func mergeTwoListsV1_2(list1 *ListNode, list2 *ListNode) *ListNode {
+	if list1 == nil {
+		return list2
+	} else if list2 == nil {
+		return list1
+	}
+
+	if list1.Val < list2.Val {
+		list1.Next = mergeTwoListsV1_2(list1.Next, list2)
+		return list1
+	} else {
+		list2.Next = mergeTwoListsV1_2(list1, list2.Next)
+		return list2
+	}
+}
+
+func mergeTwoListsV2_1(list1 *ListNode, list2 *ListNode) *ListNode {
+	var head = &ListNode{}
+	curr := head
+
+	for list1 != nil && list2 != nil {
+		if list1.Val < list2.Val {
+			curr.Next = list1
+			list1 = list1.Next
+		} else {
+			curr.Next = list2
+			list2 = list2.Next
+		}
+		curr = curr.Next
+	}
+
+	if list1 != nil {
+		curr.Next = list1
+	} else if list2 != nil {
+		curr.Next = list2
+	}
+
+	return head.Next
+}
+
+func mergeTwoListsV2_2(list1 *ListNode, list2 *ListNode) *ListNode {
+	if list1 == nil {
+		return list2
+	} else if list2 == nil {
+		return list1
+	}
+
+	head := list1
+	if list1.Val > list2.Val {
+		head = list2
+		list2 = list2.Next
+	} else {
+		list1 = list1.Next
+	}
+	curr := head
+
+	for list1 != nil && list2 != nil {
+		if list1.Val < list2.Val {
+			curr.Next = list1
+			list1 = list1.Next
+		} else {
+			curr.Next = list2
+			list2 = list2.Next
+		}
+		curr = curr.Next
+	}
+
+	if list1 != nil {
+		curr.Next = list1
+	} else if list2 != nil {
+		curr.Next = list2
+	}
+
+	return head
+}
+
+```

reverse-linked-list/invidam.go.md

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+# Intuition
+두 변수간의 Swap이 떠올랐다.
+
+# Approach
+<!-- Describe your approach to solving the problem. -->
+- 링크드 리스트의 "언제나 자신의 값과 다음 노드의 주소를 유지한다"는 특성이 재귀 함수를 유용하게 쓸 수 있다고 판단했다.
+# Complexity
+- Time complexity: $$O(n)$$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(n)$$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+
+(N은 링크드리스트의 길이)
+*/
+
+# Code
+- V1은 재귀호출, V2는 반복문 활용을 한 경우이다.
+- V3은 V1을 개선하여, 하나의 함수만으로 간결하게 해결했다.
+```
+
+func reverseListV3(head *ListNode) *ListNode {
+	if head == nil || head.Next == nil {
+		return head
+	}
+
+	newHead := reverseList(head.Next)
+
+	head.Next.Next = head
+	head.Next = nil
+
+	return newHead
+}
+
+func reverseListV2(head *ListNode) *ListNode {
+	var prev *ListNode
+	curr := head
+	for curr != nil {
+		next := curr.Next
+		curr.Next = prev
+		prev = curr
+		curr = next
+	}
+	return prev
+}
+
+func reverseNodeAllV1(head *ListNode, prev *ListNode) *ListNode {
+	if head == nil {
+		return prev
+	}
+	next := head.Next
+	head.Next = prev
+	return reverseNodeAllV1(next, head)
+}
+func reverseListV1(head *ListNode) *ListNode {
+	return reverseNodeAllV1(head, nil)
+}
+```
+
+# Remind
+재귀 함수의 특성(자기 자신을 호출한 이후 이전의 코드를 실행할 수 있다는 점)을 활용해 간결한 코딩이 가능하다.