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
/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Solution {
public:
vector<TreeNode*> doConvert(TreeNode* pRootOfTree) {
vector<TreeNode*> res();
if (pRootOfTree->left == nullptr) {
res[] = pRootOfTree;
} else {
vector<TreeNode*> leftRes = doConvert(pRootOfTree->left);
res[] = leftRes[];
leftRes[]->right = pRootOfTree;
pRootOfTree->left = leftRes[];
} if (pRootOfTree->right == nullptr) {
res[] = pRootOfTree;
} else {
vector<TreeNode*> rightRes = doConvert(pRootOfTree->right);
res[] = rightRes[];
pRootOfTree->right = rightRes[];
rightRes[]->left = pRootOfTree;
}
return res;
} TreeNode* Convert(TreeNode* pRootOfTree) {
if(pRootOfTree == nullptr) return nullptr;
vector<TreeNode* >res = doConvert(pRootOfTree);
return res[];
} };
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() - ) {
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 + , res);
swap(str[k], str[i]);
}
}
}
public:
vector<string> Permutation(string str) {
vector<string> res;
dfs(str, , 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 == ) return ;
int num = numbers[],count = ;
for(int i=;i<n;i++){
if(numbers[i] == num) count++;
else count--;
if(count == ){
num = numbers[i];
count = ;
}
}
//verify
count = ;
for(int i=;i<n;i++){
if(numbers[i] == num) count++;
}
if(count* > n ) return num;
else return ; // 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。
class Solution1 {
private:
int partition(vector<int>& input, int left, int right) {
if (left >= right)
return left;
int pivotVal = input[left];
while (left < right) {
while (left < right && input[right] >= pivotVal) {
right--;
}
input[left] = input[right];
while (left < right && input[left] <= pivotVal) {
left++;
}
input[right] = input[left];
}
input[left] = pivotVal;
return left;
}
public:
vector<int> GetLeastNumbers_Solution(vector<int>& input, int k) {
vector<int> res;
int n = input.size();
int index = partition(input, , n - );
while (index != k - ) {
if (index > k - ) {
index = partition(input, , index - );
} else {
index = partition(input, index + , n - );
}
}
for (int i = ; i < k; i++) {
res.push_back(input[i]);
}
return res;
}
};
class Solution2 {
public:
vector<int> GetLeastNumbers_Solution(vector<int>& input, int k) {
int n = input.size();
if (n == || n < k)
return vector<int>();
multiset<int, greater<int>> leastNums(input.begin(), input.begin() + k);
for (int i = k; i < n; i++) {
if (input[i] < *leastNums.begin()) {
leastNums.erase(leastNums.begin());
leastNums.insert(input[i]);
}
}
return vector<int>(leastNums.begin(), leastNums.end());
}
};
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 == ) return ;
int* dp = new int[n];
dp[] = array[];
int maxSum = dp[];
for (int i = ; i < n; i++) {
dp[i] = max(dp[i - ] + array[i], array[i]);
if (dp[i] > maxSum) {
maxSum = dp[i];
}
}
delete [] dp;
return maxSum;
}
};