标签：size sts string swa page wap 数组 超过一半 visit
26.输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向
class Solution { public: //0:left 1: right TreeNode* doConvert(TreeNode* pRootOfTree, int leftOrRight) { if (pRootOfTree == nullptr) return nullptr; TreeNode* pleft = doConvert(pRootOfTree->left, 0); TreeNode* pright = doConvert(pRootOfTree->right, 1); if (pleft) pleft->right = pRootOfTree; if (pright) pright->left = pRootOfTree; pRootOfTree->left = pleft; pRootOfTree->right = pright; //left,返回最右边的节点 if (leftOrRight == 0) { while (pRootOfTree->right) { pRootOfTree = pRootOfTree->right; } } //right,返回最左边的节点 if (leftOrRight == 1) { while (pRootOfTree->left) { pRootOfTree = pRootOfTree->left; } } return pRootOfTree; } public: TreeNode* Convert(TreeNode* pRootOfTree) { if (pRootOfTree == nullptr) return nullptr; TreeNode* pleft = doConvert(pRootOfTree->left, 0); TreeNode* pright = doConvert(pRootOfTree->right, 1); if (pleft) pleft->right = pRootOfTree; if (pright) pright->left = pRootOfTree; pRootOfTree->left = pleft; pRootOfTree->right = pright; //right,返回最左边的节点 while (pRootOfTree->left) { pRootOfTree = pRootOfTree->left; } return pRootOfTree; } };
27.输入一个字符串,按字典序打印出该字符串中字符的所有排列。例如输入字符串abc,则打印出由字符a,b,c所能排列出来的所有字符串abc,acb,bac,bca,cab和cba
class Solution { private: void dfs(string str, int k, vector<string>& res) { if (k == str.size() - 1) { res.push_back(str); return; } unordered_set<int> visited; sort(str.begin()+k,str.end()); for (size_t i = k; i < str.size(); i++) { if (visited.find(str[i]) == visited.end()) { visited.insert(str[i]); swap(str[k], str[i]); dfs(str, k + 1, res); swap(str[k], str[i]); } } } public: vector<string> Permutation(string str) { vector<string> res; dfs(str, 0, res); return res; } };
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。例如输入一个长度为9的数组{1,2,3,2,2,2,5,4,2}。由于数字2在数组中出现了5次，超过数组长度的一半，因此输出2。如果不存在则输出0。
class Solution { public: int MoreThanHalfNum_Solution(vector<int> numbers) { int n = numbers.size(); if(n == 0) return 0; int num = numbers[0],count = 1; for(int i=1;i<n;i++){ if(numbers[i] == num) count++; else count--; if(count == 0){ num = numbers[i]; count = 1; } } //verify count = 0; for(int i=0;i<n;i++){ if(numbers[i] == num) count++; } if(count*2 > n ) return num; else return 0; // not exist } };
29.输入n个整数，找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字，则最小的4个数字是1,2,3,4,。solution1在牛客网上超时，solution2能正常通过。
1 class Solution1 { 2 private: 3 int partition(vector<int>& input, int left, int right) { 4 if (left >= right) 5 return left; 6 int pivotVal = input[left]; 7 while (left < right) { 8 while (left < right && input[right] >= pivotVal) { 9 right--; 10 } 11 input[left] = input[right]; 12 while (left < right && input[left] <= pivotVal) { 13 left++; 14 } 15 input[right] = input[left]; 16 } 17 input[left] = pivotVal; 18 return left; 19 } 20 public: 21 vector<int> GetLeastNumbers_Solution(vector<int>& input, int k) { 22 vector<int> res; 23 int n = input.size(); 24 25 int index = partition(input, 0, n - 1); 26 while (index != k - 1) { 27 if (index > k - 1) { 28 index = partition(input, 0, index - 1); 29 } else { 30 index = partition(input, index + 1, n - 1); 31 } 32 } 33 for (int i = 0; i < k; i++) { 34 res.push_back(input[i]); 35 } 36 return res; 37 } 38 }; 39 40 class Solution2 { 41 public: 42 vector<int> GetLeastNumbers_Solution(vector<int>& input, int k) { 43 int n = input.size(); 44 if (n == 0 || n < k) 45 return vector<int>(); 46 multiset<int, greater<int>> leastNums(input.begin(), input.begin() + k); 47 48 for (int i = k; i < n; i++) { 49 if (input[i] < *leastNums.begin()) { 50 leastNums.erase(leastNums.begin()); 51 leastNums.insert(input[i]); 52 } 53 } 54 return vector<int>(leastNums.begin(), leastNums.end()); 55 } 56 };
30HZ偶尔会拿些专业问题来忽悠那些非计算机专业的同学。今天测试组开完会后,他又发话了:在古老的一维模式识别中,常常需要计算连续子向量的最大和,当向量全为正数的时候,问题很好解决。但是,如果向量中包含负数,是否应该包含某个负数,并期望旁边的正数会弥补它呢？例如:{6,-3,-2,7,-15,1,2,2},连续子向量的最大和为8(从第0个开始,到第3个为止)。你会不会被他忽悠住？(子向量的长度至少是1)
class Solution { public: int FindGreatestSumOfSubArray(vector<int>& array) { int n = array.size(); if (n == 0) return 0; int* dp = new int[n]; dp[0] = array[0]; int maxSum = dp[0]; for (int i = 1; i < n; i++) { dp[i] = max(dp[i - 1] + array[i], array[i]); if (dp[i] > maxSum) { maxSum = dp[i]; } } delete [] dp; return maxSum; } };
标签：size sts string swa page wap 数组 超过一半 visit
原文地址：http://www.cnblogs.com/wxquare/p/6641552.html
