LeetCode的medium题集合(C++实现)十七

1 Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.You should preserve the original relative order of the nodes in each of the two partitions.For example:

Given 1->4->3->2->5->2 and x = 3,

return 1->2->2->4->3->5.

为了方便对链表第一个节点的操作,这里要用到指向第一个节点的头指针,从头指针开始遍历,我们要用一个变量记录第一次出现节点中元素值大于目标值的位置,继续向后遍历,将小于目标的节点移动变量记录的位置之前。

ListNode* partition(ListNode* head, int x) {
        if(head==NULL) return NULL;
        ListNode* result=new ListNode(0);
        result->next=head;
        ListNode* pre=result;
        ListNode* premid;
        ListNode* cur=head;
        ListNode* mid;
        bool flag=true;  //用来标记找到第一个大于目标值的节点
        while(pre->next)
        {
            if(pre->next->val<x)
            {
                if(flag)
                {
                    pre=cur;
                    cur=cur->next;
                }
                else
                {
                    pre->next=cur->next;
                    ListNode* reg=cur;
                    cur=cur->next;
                    premid->next=reg;
                    reg->next=mid;
                    premid=premid->next;
                }
            }
            else
            {
               if(flag)
               {
                   premid=pre;
                   mid=pre->next;
                   flag=false;
                   pre=cur;
                   cur=cur->next;
               }
               else
               {
                   pre=cur;
                    cur=cur->next;
               }
            }
        }
        return result->next;
    }

Gray Code

The gray code is a binary numeral system where two successive values differ in only one bit.Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.

For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:

00 - 0

01 - 1

11 - 3

10 - 2.

可以从对应的n位二进制码字中直接得到n位格雷码码字,步骤如下:

(1)对n位二进制的码字,从右到左,以0到n-1编号

(2) 如果二进制码字的第i位和i+1位相同,则对应的格雷码的第i位为0,否则为1(当i+1=n时,二进制码字的第n位被认为是0,即第n-1位不变)。用公式表示:Gi=Bi?Bi+1,(n?1≥i≥0)。

 vector<int> grayCode(int n) {
        vector<int> result;
        int len=1<<n;
        for(int i=0;i<len;i++)
        {
            result.push_back((i>>1)^i);
        }
        return result;
    }
时间: 2024-10-07 23:56:20

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