解題思路：將一個鏈表分成三個鏈表，最后重新連接三個鏈表 （1）分成三個鏈表的過程是一致的：以左小鏈表為例依次遍歷原鏈表的每一個節點cur，當cur->data<pivot時，把cur取出， 添加到左小鏈表末尾，直到遍歷完整個原鏈表為止。 （2）生成左小、中相等、右大鏈表后，怎么連接三個鏈表并返回頭結點是難點！（因為不知道三個鏈表是否為空） #include<iostream> using namespace std;typedef struct TNode {int data;struct TNode* next; }TNode;TNode* createL(int n) //創建鏈表 {TNode* head = NULL;TNode* tial = NULL;for (int i = 0; i < n; i++){TNode* node = new TNode;node->next = NULL;cout << "請輸入";cin >> node->data;if (head == NULL){head = node;tial = head;}else{tial->next = node;tial = node;}}tial->next = NULL;return head; }void travel(TNode* head) {cout << "輸出結果：" << endl;TNode* p = head;while (p != NULL){cout << p->data << endl;p = p->next;} }TNode* listPartition(TNode* head, int pivot) //鏈表劃分 {TNode* sh = NULL;TNode* st = NULL;TNode* eh = NULL;TNode* et = NULL;TNode* bh = NULL;TNode* bt = NULL;//依次遍歷每一個元素TNode* cur = head;TNode* s = NULL; //輔助變量，用于保存cur節點的下一個節點，防止斷鏈while (cur != NULL){s = cur->next;cur->next = NULL;//將當前結點cur的指針域設為空，保證每個鏈表的新增節點為最后一個節點if (cur->data < pivot){if (sh == NULL){sh = cur;st = cur;}else{st->next = cur;st = cur;}}else if (cur->data == pivot){if (eh == NULL){eh = cur;et = cur;}else{et->next = cur;et = cur;}}else if (cur->data > pivot){if (bh == NULL){bh = cur;bt = cur;}else{bt->next = cur;bt = cur;}}cur = s;}//難點：將三個不知道是否為空鏈表的鏈表進行鏈接成一條鏈表，返回最終的頭結點 //依次從左到右假設每個鏈表不為空時，尾節點指針域的指向if (st != NULL){st->next = (eh != NULL) ? eh : bh;}if (et != NULL){et->next = bh;}return st != NULL ? sh : et != NULL ? eh : bh; } int main() {TNode* head = NULL;head = createL(7);travel(head);head = listPartition(head, 3);travel(head);return 0; }

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