String declaration in Java

http://blog.sina.cn/dpool/blog/s/blog_5b9b4abe01017b4q.html

Java consumer producers problem using mutex

http://java67.blogspot.com/2012/12/producer-consumer-problem-with-wait-and-notify-example.html?m=1

156. Binary Tree Upside Down ---M ~~~~做了3遍的, 竟然忘了~~~~

Given the root of a binary tree, turn the tree upside down and return the new root.

You can turn a binary tree upside down with the following steps:

1. The original left child becomes the new root.
2. The original root becomes the new right child.
3. The original right child becomes the new left child.

The mentioned steps are done level by level, it is guaranteed that every node in the given tree has either 0 or 2 children.

 

Example 1:

Input: root = [1,2,3,4,5]
Output: [4,5,2,null,null,3,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

 

Constraints:

• The number of nodes in the tree will be in the range [0, 10].
• 1 <= Node.val <= 10
• Every node has either 0 or 2 children.

A:

这个题目描述的很烂。

应该是： binary tree upside down along left most edge

Iterative bottom up.
就是先做左子树，再连接修改该节点。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* upsideDownBinaryTree(TreeNode* root) {
        if(not root){ // if nullptr , or no child
            return root;
        }
        stack<TreeNode*> S;
        TreeNode *runner = root;
        while(runner->left ){
            S.push(runner);
            runner = runner->left;
        }
        TreeNode * newRoot = runner;
        while(not S.empty()){
            auto p = S.top();
            S.pop();
            runner->right = p;
            runner->left = p->right;
            p->left = nullptr;
            p->right = nullptr;
            runner = p;
        }
        return newRoot;
    }
};

错误： 

没有先把  p.right = nullptr； 

这个题比较tricky的点就是,要事先找到return 的newRoot.




------------Recursive approach--------先找到 返回TreeNode， 这样就不需要在helper()中管了--------

public class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if(root == null)
            return null;
        TreeNode leftMost = root;
        while( leftMost.left != null)
            leftMost = leftMost.left;
            
        helper(root);
        return leftMost;
    }
    private void helper(TreeNode  root){
        if(root.left == null)
            return;
        upsideDownBinaryTree(root.left);
        
        TreeNode newRoot = root.left;
        newRoot.left = root.right;
        newRoot.right = root;
        
        root.left = null;
        root.right = null;
    }
}

这里利用了一个性质，就是指针指向的Node只做了child 指针的更改操作。原本的Object 位置并没有更改。

Mistakes：
1: base case中，忘了把root.left==null的情况加进来（否则会有null pointer的Exception)

public class Solution {
    public TreeNode UpsideDownBinaryTree(TreeNode root) {
        if(root == null || root.left == null)
            return root;
        TreeNode newRoot = UpsideDownBinaryTree(root.left);
        
        TreeNode r= root.right;
        root.left =null;
        root.right = null;
        TreeNode runner = newRoot;
        while(runner!=null && runner.right!=null){
            runner = runner.right;
        }
        if(runner!=null){
            runner.left = r;
            runner.right = root;
        }
        return newRoot;
    }
}

155. Min Stack (Medium)

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the MinStack class:

• MinStack() initializes the stack object.
• void push(int val) pushes the element val onto the stack.
• void pop() removes the element on the top of the stack.
• int top() gets the top element of the stack.
• int getMin() retrieves the minimum element in the stack.

You must implement a solution with O(1) time complexity for each function.

 

Example 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

 

Constraints:

• -231 <= val <= 231 - 1
• Methods pop, top and getMin operations will always be called on non-empty stacks.
• At most 3 * 104 calls will be made to push, pop, top, and getMin.

A:

思想很简单。
后来出现memory out of limit, minStack中就没有存多余的最小值。

class MinStack {
public:
    /** initialize your data structure here. */
    vector<int> stack;
    vector<int> minStack;
    MinStack() {
        
    }
    
    void push(int x) {
        stack.push_back(x);
        if(minStack.empty() || minStack.back() > x)
            minStack.push_back(x);
        else
            minStack.push_back(minStack.back());
    }
    
    void pop() {
        stack.pop_back();
        minStack.pop_back();
    }
    
    int top() {
        return stack.back();
    }
    
    int getMin() {
        return minStack.back();
    }
};

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack* obj = new MinStack();
 * obj->push(x);
 * obj->pop();
 * int param_3 = obj->top();
 * int param_4 = obj->getMin();
 */

Learned:

1;  minStack中，重复的最小值也要存。
2:即使Stack<Integer>的返回值是一个对象，因此要用.equals()来比较大小