sol 92 and 206, reverseBetween and reverseList, include the test project

Author: ioMayday

c_code/92_206_reverseBetween_test.c

@@ -0,0 +1,87 @@
+// 参考博文：
+// https://labuladong.gitbook.io/algo/shu-ju-jie-gou-xi-lie/shou-ba-shou-shua-lian-biao-ti-mu-xun-lian-di-gui-si-wei/di-gui-fan-zhuan-lian-biao-de-yi-bu-fen
+
+// Definition for singly-linked list.
+struct ListNode {
+    int val;
+    struct ListNode *next;
+};
+
+// 先实现整个链表反转 leetcode题206
+struct ListNode* reverseList(struct ListNode* head){
+    if (head == NULL) return NULL; // 注意，要先判head是否为空
+    if (head->next == NULL) return head;
+
+    struct ListNode* last = reverseList(head->next);
+    head->next->next = head;
+    head->next = NULL;
+    return last;
+}
+
+// 再实现前n个元素反转
+struct ListNode* succesor = NULL;
+struct ListNode* reverseN(struct ListNode* head, int n)
+{
+    if (n == 1) {
+        succesor = head->next;
+        return head;
+    }
+
+    struct ListNode* last = reverseN(head->next, n - 1); //只迭代展开一个，不要压栈出栈
+    head->next->next = head;
+    head->next = succesor;
+    return last;
+}
+
+// 再实现从第m到n的两侧均闭区间的反转
+struct ListNode* reverseBetween(struct ListNode* head, int left, int right){
+    if (head == NULL) return NULL;  // left/right值题目已做限定，默认是正确的
+
+    // 如果left=1,直接是起点
+    if (left == 1) {
+        return reverseN(head, right);  // left移至1时，right大小等于n个数
+    }
+    // 否则，移动到对应起点
+    head->next = reverseBetween(head->next, left - 1, right - 1); // 返回反转片段新的地点
+    return head;
+}
+
+
+
+//主函数调用
+int main()
+{
+    struct ListNode head, p2, p3, p4, p5;
+    head.val = 1;
+    head.next = &p2;
+    p2.val = 2;
+    p2.next = &p3;
+    p3.val = 3;
+    p3.next = &p4;
+    p4.val = 4;
+    p4.next = &p5;
+    p5.val = 5;
+    p5.next = NULL;
+
+    // 初始化后遍历显示
+    struct ListNode *cur, *pre, *nxt;
+    cur = &head;
+    while (cur != NULL) {
+        printf("%d ",cur->val);
+        cur = cur->next;
+    }
+    printf("\n");
+
+    // 反转后遍历显示
+//    struct ListNode* newHead = reverseN(&head, 3);
+    struct ListNode* newHead = reverseBetween(&head, 2, 5);
+    cur = newHead;
+    while (cur != NULL) {
+        printf("%d ",cur->val);
+        cur = cur->next;
+    }
+    printf("\n");
+	
+	return 0;
+
+}

c_code/92_reverseBetween.c

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+struct ListNode* reverseBetween(struct ListNode* head, int left, int right){
+    if (head == NULL) return NULL;  // left/right值题目已做限定，默认是正确的
+
+    // 如果left=1,直接是起点
+    if (left == 1) {
+        return reverseN(head, right);  // left移至1时，right大小等于n个数
+    }
+    // 否则，移动到对应起点
+    head->next = reverseBetween(head->next, left - 1, right - 1); // 返回反转片段新的地点
+    return head;
+}