April Day 8 Challenge Done

problems/876/main.c

@@ -0,0 +1,41 @@
+#include <stdio.h>
+
+struct ListNode {
+    int val;
+    struct ListNode *next;
+};
+
+struct ListNode* middleNode(struct ListNode*);
+struct ListNode* middleNode(struct ListNode* head){
+    struct ListNode* ind = head;
+    struct ListNode* middle = head;
+    int on = 0;
+    while (ind != NULL) {
+        ind = ind->next;
+        if (on == 0) {
+            on = 1;
+        } else {
+            on = 0;
+            middle = middle->next;
+        }
+    }
+    return middle;
+}
+
+int main() {
+    printf("Expected: [4, 5, 6]\n");
+    printf("Got:");
+    struct ListNode l6 = { 6, NULL };
+    struct ListNode l5 = { 5, &l6 };
+    struct ListNode l4 = { 4, &l5 };
+    struct ListNode l3 = { 3, &l4 };
+    struct ListNode l2 = { 2, &l3 };
+    struct ListNode linkedList = { 1, &l2 }; 
+    struct ListNode* middle = middleNode(&linkedList);
+    while (middle != NULL) {
+        printf(" %d ", middle->val);
+        middle = middle->next;
+    }
+    printf("\n");
+    return 0;
+}

problems/876/problem.txt

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+Given a non-empty, singly linked list with head node head, return a middle node of linked list.
+
+If there are two middle nodes, return the second middle node.
+
+ 
+
+Example 1:
+
+Input: [1,2,3,4,5]
+Output: Node 3 from this list (Serialization: [3,4,5])
+The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).
+Note that we returned a ListNode object ans, such that:
+ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
+
+Example 2:
+
+Input: [1,2,3,4,5,6]
+Output: Node 4 from this list (Serialization: [4,5,6])
+Since the list has two middle nodes with values 3 and 4, we return the second one.
+
+ 
+
+Note:
+
+    The number of nodes in the given list will be between 1 and 100.
+

problems/876/run.sh

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+#!/bin/bash
+
+gcc -o main main.c
+./main
+rm main