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求问以下代码的问题，在本地ide上跑了几组样例能过😣😣😣 class Solution { public: ListNode* ReverseList(ListNode* pHead) { ListNode* p = pHead; stack<int> s; while(p->next != NULL){ s.push(p->val); p = p->next; } ListNode* new_p = p; while(s.size()>0){ ListNode *node = new ListNode(s.top()); new_p->next = node; s.pop(); new_p = new_p->next; } return p; } };
class Solution { public: bool Find(int target, vector<vector<int> > array) { int i=0,j=0 ,low ,high; int m = array.size() , n = array[0].size() ; for(i = 0 ; i < m ; i ++ ) { low = 0; high = n-1; int mid; while( low < high ) { mid = ( low + high )/2; if( array[i][mid] >= target ) high = mid; else if(array[i][mid] < target)low = mid+1; } if(mid <n && array[i][low] == target)return true; } return false; } };
public class Solution { public String replaceSpace(StringBuffer str) { return str.toString().replaceAll(" ","%20"); } } 要习惯使用Java自带函数
class Solution { public: vector<int> printListFromTailToHead(ListNode* head) { unsigned long length=0; ListNode * pre= nullptr; while (head!= nullptr){ ListNode * temp=head; head=head->next; temp->next=pre; pre=temp; ++length; } vector<int> result(length); unsigned long index=0; head=pre; while (head!= nullptr){ result[index++]=head->val; head=head->next; } return result; } }; 遍历的同时反转链表，同时记录链表的长度。最后再遍历一遍放入数组。时间复杂度2*n
//java //从后往前找，如果找到比数组第一个数大的就停下来 public class Solution { public int minNumberInRotateArray(int [] array) { int length = array.length; if(length==0) return 0; int index=length-1; while(array[index]-array[0]<=0){ index--; } return array[index+1]; } }
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