Chap5: question 35

35. 第一个只出现一次的字符

char firtNotRepeat(char *s)
{
    if(s == NULL) return 0;
    int i = 0;
    while(s[i] != '')  record[s[i++]] ^= 1;
    i = 0;
    while(!record[s[i]]) ++i;
    return s[i];
}

36.数组中的逆序对个数 （归并排序解法）

#include <iostream>
using namespace std;
void inversePairsCore(int data[], int copy[], int low, int high, int& count)
{
    if(low == high) return;

    int mid = (low + high) / 2;
    inversePairsCore(data, copy, low, mid, count);
    inversePairsCore(data, copy, mid+1, high, count);

    int k = high, tag_mid = mid, tag_high = high;
    while(low <= mid && tag_mid +1 <= high)
    {
        if(data[mid] > data[high])
        {
            copy[k--] = data[mid--];
            count += high - tag_mid;
        }
        else if(data[mid] < data[high])
        {
            copy[k--] = data[high--];
        }
        else 
            copy[k--] = data[high--];
    }
    while(low <= mid) copy[k--] = data[mid--];
    while(tag_mid+1 <= high) copy[k--] = data[high--];

    for(k = low; k <= tag_high; ++k)
        data[k] = copy[k];
}
int inversePairs(int data[], int length)
{
    if(data == NULL && length < 1) return 0;

    int *copy = new int[length];
    int count = 0;
    inversePairsCore(data, copy, 0, length-1, count);

    delete[] copy;
    return count;
}

int main()
{
    int data[] = {4, 4, 4, 3, 3};
    printf("%d
", inversePairs(data, sizeof(data)/4));
    return 0;
}

37.  两个链表的第一个公共结点

int getLength(ListNode *pHead)
{
    if(pHead == NULL) return 0;
    int len = 0;
    ListNode *p = pHead;
    while(p != NULL)
    {
        p = p->next;
        len ++;
    }
    return len;
}
ListNode* firstNode(ListNode *pHead1, ListNode* pHead2)
{
    if(pHead1 == NULL || pHead2 == NULL) return NULL;
    int len1 = getLength(pHead1);
    int len2 = getLength(pHead2);
    if(len1 < len2) return firstNode(pHead2, pHead1);
    int k = len1 - len2;
    ListNode *p1 = pHead1, *p2 = pHead2;
    while(k > 0) 
    {
        p1 = p1->next;
        --k;
    }
    while(p1 != NULL && p2 != NULL && p1 != p2)
    {
        p1 = p1->next;
        p2 = p2->next;
    }
    if(p1 == NULL || p2 == NULL) return NULL;
    return p1;
}