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剑指offer题25:复制一个随机链表
RondomListNode* RandomLinkCopy(RondomListNode *head)
{
if (!head)return NULL;//头结点是空直接返回
RondomListNode *cur = head, *next = head->next;//由①②③变成①①②②③③,但复制出来的结点的随机指针都指向空
while (cur){
RondomListNode *tmp = new RondomListNode(cur->val);
cur->next = tmp; tmp->next = next;
cur = next; if (next)next = next->next;
}
cur = head; next = cur->next;//将随机指针指向正确的位置
while (cur){
next->rand = cur->rand ? cur->rand->next : NULL;
cur = next->next; if (cur)next = cur->next;
}
cur = head; next = cur->next;//将新旧链表分离开
RondomListNode *tmp = next;
while (cur){
cur->next = next->next; cur = next->next;
if (cur){ next->next = cur->next; next = cur->next; }
else next->next = NULL;
}
return tmp;//返回复制出来的新链表的头结点
}
剑指offer题26:输入一棵二叉搜索树,将该二叉树转换成一个排序的双向链表。要求不能创建新的结点,只能调整树中结点的指针
TreeNode *_TreeToLink(TreeNode *root)
{
if (!root)return NULL;//递归返回条件
TreeNode *left = _TreeToLink(root->left);//left是左边单链表的最左结点
TreeNode *right = _TreeToLink(root->right);//right是右边单链表的最左结点
TreeNode *left_to_right;//左边单链表的最右结点
left_to_right = left ? left->left : NULL;
TreeNode *right_to_right;//右边单链表的最右结点
right_to_right = right ? right->left : NULL;
root->left = left_to_right;//根的左孩子
if (left_to_right)left_to_right->right = root;
root->right = right;//根的右孩子
if (right)right->left = root;
if (!left)left = root;
left->left = right ? right_to_right : root;
return left;
}
TreeNode *TreeToLink(TreeNode *root)//递归思路
{
if (!root)return NULL;
TreeNode *head = _TreeToLink(root);
head->left = NULL;
return head;
}
剑指offer题28:数组中有一个数字出现的次数超过数组长度的一半,请找出这个数字,ps:如果不存在返回0.
方法一:排序 方法二:map<K,V> 方法三:阵地攻守思想,先让第一个元素作为第一个士兵来守阵地,count=1;
遇到相同的元素++count,否则--count;当count=0时,又以新的i值作为守阵地的士兵,直到最后的士兵,那即是所求元素
方法四:用堆找出一般元素,返回堆顶
int MoreThanHalfNum(vector<int> arr)//方法三
{
if (arr.empty())return 0;
int count = 1, soldier = arr[0];
for (int i = 1; i < arr.size(); ++i)
{
arr[i] == soldier ? ++count : --count;
if (count == 0)
{
soldier = arr[i];
count = 1;
}
}
count = 0;
for (int i = 0; i < arr.size(); ++i)
{
if (arr[i] == soldier)
++count;
}
return count > arr.size() / 2 ? soldier : 0;
}
剑指offer题29:输入n个整数,找出其中最小的k个数
#include<queue>
vector<int> GetLeastNumbers(vector<int> input, int k)
{
if (input.size() < k || k <= 0)return vector<int>();//特别注意k的取值
priority_queue<int> pq;//找最小的前k个数,建大根堆 (孩子 less 双亲)
for (int i = 0; i < input.size(); ++i)
{
if (i < k)
pq.push(input[i]);
else if (pq.top() > input[i])
{
pq.pop();
pq.push(input[i]);
}
}
vector<int> arr(k); int index = k - 1;
while (!pq.empty())
{
arr[index--] = pq.top();
pq.pop();
}
return arr;
}
剑指offer题30:输入一个数组,求该数组的最大子数组和是多少。ps,该数组长度最少为1.
int FindGreatestSumOfSubArray(vector<int> arr)
{
int max = arr[0], val = arr[0];
for (size_t i = 1; i < arr.size(); ++i)
{
val = val < 0 ? 0 : val;//确定val是保留原值还是置零
val += arr[i];
if (val > max)max = val;//max代表原数组[0,i]范围内的最大子数组加和
}
return max;
}