26. 順時針打印矩陣
因爲弄反了行列，所以因爲溢出折騰了好一會。做之前感覺遞歸會更方便，現在還是覺得循環更簡單些。
class Solution {
public:
void printM(vector<vector<int> > matrix,int rowbegin,int rowend,int colbegin,int colend,vector<int>& result){
if(rowbegin>rowend || colbegin>colend)
return;
int i=colbegin,j=rowbegin;
if(rowbegin==rowend){
while(i<=colend) {result.push_back(matrix[i][j]);i++;}
return;
}
else if(colbegin==colend){
while(j<=rowend) {result.push_back(matrix[i][j]);j++;}
return;
}
while(j<rowend) {result.push_back(matrix[i][j]);j++;}
while(i<colend) {result.push_back(matrix[i][j]);i++;}
while(j>rowbegin) {result.push_back(matrix[i][j]);j--;}
while(i>colbegin) {result.push_back(matrix[i][j]);i--;}
if(rowbegin<rowend && colbegin<colend)
printM(matrix,rowbegin+1,rowend-1,colbegin+1,colend-1,result);
return;
}
vector<int> printMatrix(vector<vector<int> > matrix) {
vector<int> result;
if(matrix.size()==0) return result;
if(matrix[0].size()==0) return result;
int rowend=matrix[0].size()-1;
int colend=matrix.size()-1;
int rowbegin=0,colbegin=0;
if(rowbegin<=rowend && colbegin<=colend)
printM(matrix,rowbegin,rowend,colbegin,colend,result);
return result;
}
};
27. 包含min函數的棧
class Solution {
public:
stack<int> stack1,stack2;
void push(int value) {
stack1.push(value);
if(stack2.empty()) stack2.push(value);
else{
if(value<=stack2.top()) stack2.push(value);
}
}
void pop() {
if(stack1.top()==stack2.top()) stack2.pop();
stack1.pop();
}
int top() {
return stack1.top();
}
int min() {
return stack2.top();
}
};
28. 棧的壓入、彈出序列
建立輔助棧，模擬棧的壓入彈出過程看是否能全彈出.
class Solution {
public:
bool IsPopOrder(vector<int> pushV,vector<int> popV) {
if(pushV.size()==0 || popV.size()==0) return false;
stack<int> stk;
int i=0;
for(int j=0;j<pushV.size() && i<popV.size();j++){
stk.push(pushV[j]);
while(i<popV.size() && !stk.empty() && stk.top()==popV[i]){
stk.pop();
i++;
}
}
if(stk.empty()) return true;
return false;
}
};
29. 打印二叉樹
29.1 從上到下打印二叉樹
使用隊列，先進先出，按順序保留每個子節點。
class Solution {
public:
vector<int> PrintFromTopToBottom(TreeNode* root) {
vector<int> print;
if(root==NULL) return print;
queue<TreeNode*> que;
que.push(root);
while(!que.empty()){
root=que.front();
que.pop();
if(root==NULL) continue;
print.push_back(root->val);
que.push(root->left);
que.push(root->right);
}
return print;
}
};
29.2. 分行從上到下打印二叉樹
使用一個int值計數，記下一層節點個數countnext，當前節點個數count會在遍歷完當前層之後重新賦值爲countnext，開始遍歷下一層。
class Solution {
public:
vector<vector<int> > Print(TreeNode* pRoot) {
vector<vector<int>> result;
vector<int> col;
if(pRoot==NULL) return result;
queue<TreeNode*> que;
que.push(pRoot);
int count=1,countnext=0;
while(!que.empty()){
pRoot=que.front();
if(pRoot->left!=NULL){
que.push(pRoot->left);
countnext++;
}
if(pRoot->right!=NULL){
que.push(pRoot->right);
countnext++;
}
que.pop();
col.push_back(pRoot->val);
count--;
if(count==0){
result.push_back(col);
col.clear();
count=countnext;
countnext=0;
}
}
return result;
}
};
29.3 之字形打印二叉樹
class Solution {
public:
vector<vector<int> > Print(TreeNode* pRoot) {
vector<vector<int>> result;
vector<int> col;
if(pRoot==NULL) return result;
stack<TreeNode*> stknext;
stack<TreeNode*> stk;
stk.push(pRoot);
bool ceng=true;//分奇偶層；
int count=1,countnext=0;
while(!stk.empty()){
pRoot=stk.top();
//收集下一層的節點
if(pRoot->left!=NULL && ceng){
stknext.push(pRoot->left);
countnext++;
}
if(pRoot->right!=NULL){
stknext.push(pRoot->right);
countnext++;
}
if(pRoot->left!=NULL && !ceng){
stknext.push(pRoot->left);
countnext++;
}
stk.pop();
col.push_back(pRoot->val);
count--;
if(count==0){
//捋一下下一層節點的順序
result.push_back(col);
col.clear();
count=countnext;
countnext=0;
ceng=!ceng;
stk=stknext;
//因爲沒有能直接清空棧的方法，用了一個笨方法。
stack<TreeNode*> empt;
stknext=empt;
}
}
return result;
}
};
30. 二叉搜索樹的後序遍歷序列
class Solution {
public:
bool VerifySquenceOfBST(vector<int> sequence) {
//搜索樹默認中序遍歷有序，所以相當於找中序與後序的聯繫
int size=sequence.size();
if(size==0) return false;
if(size==1) return true;
vector<int> left;
vector<int> right;
int i=0;
bool leftflag=true;
while(i<size-1){
//左子樹
if(sequence[i]<sequence[size-1] && leftflag){
left.push_back(sequence[i]);
i++;
}
//右子樹
else if(sequence[i]>sequence[size-1]){
leftflag=false;
right.push_back(sequence[i]);
i++;
}
//應該在右子樹的但小於根節點，異常
else return false;
}
//最初輸入的樹爲空應 false；這裏爲了區別最初的空還是後來的空。
bool leftBST=left.size()<=1? true :VerifySquenceOfBST(left);
bool rightBST=right.size()<=1? true :VerifySquenceOfBST(right);
return leftBST && rightBST;
}
};