編程演算法 - 兩個鏈表的第一個公用結點 代碼(C)

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兩個鏈表的第一個公用結點 代碼(C)

本文地址: http://blog.csdn.net/caroline_wendy

題目: 輸入兩個鏈表, 找出它們的第一個公用結點.

計算鏈表的長度, 然後移動較長鏈表的指標, 使其到相同結點的距離的相同, 再同時移動兩個鏈表的指標, 找到相同元素.

時間複雜度: O(n)

代碼:

/* * main.cpp * *  Created on: 2014.6.12 *      Author: Spike *//*eclipse cdt, gcc 4.8.1*/#include <stdio.h>#include <stdlib.h>#include <string.h>struct ListNode {int m_nKey;ListNode* m_pNext;};size_t GetListLength (ListNode* pHead) {size_t nLength = 0;ListNode* pNode = pHead;while (pNode != NULL) {++nLength;pNode = pNode->m_pNext;}return nLength;}ListNode* FindFirstCommonNode(ListNode* pHead1, ListNode* pHead2) {size_t nLength1 = GetListLength(pHead1);size_t nLength2 = GetListLength(pHead2);int nLengthDif = nLength1 - nLength2;ListNode* pListHeadLong = pHead1;ListNode* pListHeadShort = pHead2;if (nLength2 > nLength1) {pListHeadLong = pHead2;pListHeadShort = pHead1;nLengthDif = nLength2 - nLength1;}for (int i=0; i<nLengthDif; ++i)pListHeadLong = pListHeadLong->m_pNext;while ((pListHeadLong != NULL) && (pListHeadShort != NULL)&& (pListHeadLong != pListHeadShort)) {pListHeadLong = pListHeadLong->m_pNext;pListHeadShort = pListHeadShort->m_pNext;}ListNode* pFirstCommonNode = pListHeadLong;return pFirstCommonNode;}int main(void){ListNode* pHead1 = new ListNode();ListNode* pHead1Node1 = new ListNode();ListNode* pHead1Node2 = new ListNode();ListNode* pHead1Node3 = new ListNode();ListNode* pHead1Node4 = new ListNode();pHead1->m_nKey = 1;pHead1Node1->m_nKey = 2;pHead1Node2->m_nKey = 3;pHead1Node3->m_nKey = 6;pHead1Node4->m_nKey = 7;pHead1->m_pNext = pHead1Node1;pHead1Node1->m_pNext = pHead1Node2;pHead1Node2->m_pNext = pHead1Node3;pHead1Node3->m_pNext = pHead1Node4;pHead1Node4->m_pNext = NULL;ListNode* pHead2 = new ListNode();ListNode* pHead2Node1 = new ListNode();pHead2->m_nKey = 4;pHead2Node1->m_nKey = 5;pHead2->m_pNext = pHead2Node1;pHead2Node1->m_pNext = pHead1Node3;ListNode* result = FindFirstCommonNode(pHead1, pHead2);printf("result = %d\n", result->m_nKey);    return 0;}

輸出:

result = 6