面试题37 两个链表的第一个公共结点

题目描述

输入两个链表，找出它们的第一个公共结点。

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode *pHead1, ListNode *pHead2) {
        if (pHead1 == NULL || pHead2 == NULL)
            return NULL;
        ListNode *p1 = pHead1, *p2 = pHead2;
        int l1 = 0, l2 = 0;
        while(p1){
            l1++;
            p1 = p1->next;
        }
        
        while(p2){
            l2++;
            p2 = p2->next;
        }
        
        int n = l1 - l2;
        p1 = pHead1; p2 = pHead2;
        if (l2 > l1){
            p1 = pHead2;
            p2 = pHead1;
            n = l2 - l1;
        }
        for (int i = 0; i < n; i++)
            p1 = p1->next;
        while (p1 != NULL && p2 != NULL && p1 != p2){
            p1 = p1->next;
            p2 = p2->next;
        }
        ListNode *p = p1;
        return p;
    }
};

 

class Solution {
public:
    int Merge(vector<int> &A, int p, int q, int r)
    {
        int count = 0;

        vector<int> B;
        vector<int> C;

        for (int i = p; i <= q; i++)
        {
            B.push_back(A[i]);
        }
        for (int i = q + 1; i <= r; i++)
        {
            C.push_back(A[i]);
        }

        int m = B.size() - 1, n = C.size() - 1;
        for (int i = r; i >= p; i--)
        {
            if (m == -1 && n != -1)
            {
                while (n != -1)
                {
                    A[i] = C[n];
                    i--;
                    n--;
                }
                break;
            }
            if (m != -1 && n == -1)
            {
                while (m != -1)
                {
                    A[i] = B[m];
                    i--;
                    m--;
                }
                break;
            }
            if (B[m] <= C[n])
            {
                A[i] = C[n];
                n--;
            }
            else
            {
                count += n + 1;
                A[i] = B[m];
                m--;
            }
        }
        return count;
    }

    int MergeSort(vector<int> &A, int p, int r)
    {
        if (p == r)
            return 0;
        int q = (p + r) / 2;
        int left = MergeSort(A, p, q);
        int right = MergeSort(A, q + 1, r);
        return left + right + Merge(A, p, q, r);
    }

    int InversePairs(vector<int> &data) {
        if (data.size() == 0)
            return 0;
        return MergeSort(data, 0, data.size() - 1);
    }
};