C 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2020-09-07

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

/**
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
 int findListLength(struct ListNode *pHead1)
 {
     if(pHead1==NULL)
         return 0;
     int sum=1;
     while(pHead1=pHead1->next)
         sum++;
     return sum;
     
 }
struct ListNode* walkStep(struct ListNode *pHead1,int step){
    while(step--)
    {
        pHead1=pHead1->next;
        
    }
    return pHead1;
}

struct ListNode* FindFirstCommonNode(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
   
    int len1=findListLength(pHead1);
    int len2=findListLength(pHead2);
    if(len1>len2)
        pHead1=walkStep(pHead1, len1-len2);
    else
        pHead1=walkStep(pHead1, len1-len2);
     while(pHead1 != NULL){
            if(pHead1 == pHead2) return pHead1;
            pHead1 = pHead1->next;
            pHead2 = pHead2->next;
        }
        return NULL;
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
}

C 解法, 执行用时: 2ms, 内存消耗: 340KB, 提交时间: 2021-03-21

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

/**
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
struct ListNode* FindFirstCommonNode(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    struct ListNode *p=pHead1;
    struct ListNode *q=pHead2;
    
    while(p!=q)
    {
        p = p ? p->next : pHead2;
        q = q ? q->next : pHead1;
    }
    return p;
}

C 解法, 执行用时: 2ms, 内存消耗: 340KB, 提交时间: 2020-09-06

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

/**
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
struct ListNode* FindFirstCommonNode(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    struct ListNode *p1 = pHead1, *p2 = pHead2;
    while(p1 != p2)
    {
        p1 = p1 ? p1->next : pHead2;
        p2 = p2 ? p2->next : pHead1;
    }
    return p1;
} 

C 解法, 执行用时: 2ms, 内存消耗: 340KB, 提交时间: 2020-08-30

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

/**
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
struct ListNode* FindFirstCommonNode(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    if (pHead1 == NULL || pHead2 == NULL) {
	    return NULL;
    }
    struct ListNode *p1, *p2;
    p1 = pHead1;
    p2 = pHead2;
    while (p1 != p2) {
	    p1 = (p1 == NULL ? pHead1 : p1->next);
	    p2 = (p2 == NULL ? pHead2 : p2->next);
    }
    return p1;
}

C 解法, 执行用时: 2ms, 内存消耗: 344KB, 提交时间: 2021-03-17

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

/**
 * 
 * @param pHead1 ListNode类 
 * @param pHead2 ListNode类 
 * @return ListNode类
 */
struct ListNode* FindFirstCommonNode(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    struct ListNode *res = NULL, *record_1[888], *record_2[888];
    int idx_1 = 0, idx_2 = 0;
    while(pHead1 && pHead2)
    {
        record_1[ idx_1++ ] = pHead1;
        record_2[ idx_2++ ] = pHead2;
        int i = idx_1-1, j = idx_2-1;
        while(i >= 0)
        {
            if(record_1[ i-- ] == pHead2) return pHead2;
            if(record_2[ j-- ] == pHead1) return pHead1;
        }
        pHead1 = pHead1->next;
        pHead2 = pHead2->next;
    }
    char flag = 0;
    if(pHead1){
        do{
            int j = idx_2-1;
            while(j >= 0)
                if(record_2[ j-- ] == pHead1){ res = pHead1;   flag = 1;   break; }
            if(flag) break;
        }while(pHead1 = pHead1->next);
    }
    else if(pHead2){
        do{
            int i =idx_1-1;
            while(i >= 0)
                if(record_1[ i-- ] == pHead2){ res = pHead2;   flag = 1;   break; }
            if(flag) break;
        }while(pHead2 = pHead2->next);
    }
    
    return res;
}