221. Maximal Square

動態規劃，只用一個一維陣列，這裡要注意程式碼裡的對應關係，相當於在原陣列的基礎上在前面和上面擴充了一行一列。
動態規劃的示意圖如下所示：

因此需要用prev儲存上一個dp[j-1]的值

class Solution {
    public int maximalSquare(char[][] matrix) {
        if(matrix==null || matrix.length==0)
            return 0;
        int[] dp = new int[matrix[0].length+1];
        int maxlen=0;
        int prev=0;
        for(int i=1;i<=matrix.length;i++){
            for(int j=1;j<=matrix[0].length;j++){
                int temp = dp[j];
                if(matrix[i-1][j-1] == '1')
                    dp[j] = Math.min(temp,Math.min(dp[j-1],prev)) + 1;
                else{
                    dp[j] = 0;
                }
                maxlen = Math.max(dp[j],maxlen);
                prev = temp;
            }
        }
        return maxlen * maxlen;
    }
}

222. Count Complete Tree Nodes

程式碼的核心思想在於，首先判斷樹有幾層，然後根據二分搜尋的思想，判斷最後一個節點出現的位置。用root-left-right---right的思路得到最後一個節點出現的位置。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        if(root==null) return 0;
        if(root.left == null) return 1;
        int height = 0;
        int nodeSum = 0;
        TreeNode node = root;
        while(node.left!=null){
            nodeSum += (1<<height);
            height += 1;
            node = node.left;
        }
        return nodeSum + getLastNodeSum(root,height);
    }
    
    public int getLastNodeSum(TreeNode root,int height){
        if(height==1){
            if(root.right!=null) return 2;
            else if(root.left != null) return 1;
            else return 0;
        }
        TreeNode node = root.left;
        int curHeight = 1;
        while(curHeight < height){
            node = node.right;
            curHeight += 1;
        }
        if(node == null) return getLastNodeSum(root.left,height-1);
        else
            return (1<<(height-1)) + getLastNodeSum(root.right,height-1);
        
    }
}

223. Rectangle Area

重點是計算兩個圖形的交集。

class Solution {
    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int area1 = (C-A) * (D-B);
        int area2 = (G-E) * (H-F);
        if (C <= E || G <= A || D <= F || H <= B)
            return area1 + area2;
        else
            return area1 + area2 - Math.abs(Math.max(A,E) - Math.min(C,G)) * Math.abs(Math.max(B,F)-Math.min(D,H));
    }
}

225. Implement Stack using Queues

使用佇列的功能來實現一個stack。

class MyStack {

    Queue<Integer> queue;
    
    /** Initialize your data structure here. */
    public MyStack() {
        queue = new LinkedList<Integer>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        queue.add(x);
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        for(int i=0; i<queue.size()-1; I++)
            queue.add(queue.remove());
        return queue.remove();
    }
    
    /** Get the top element. */
    public int top() {
        for(int i=0; i<queue.size()-1; I++)
            queue.add(queue.remove());
        int t = queue.remove();
        queue.add(t);
        return t;
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return queue.isEmpty();
    }
}

226. Invert Binary Tree

遞迴呼叫。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null)
            return root;
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        root.left = invertTree(root.left);
        root.right = invertTree(root.right);
        return root;
    }
}

227. Basic Calculator II

自己寫的，AC了，運用了兩個棧，一個棧儲存數字一個棧儲存運算子，第一次遍歷只需要計算乘除部分，第二次的話 需要將棧的元素反過來，舉個例子，如果經過乘除運算之後是1-1+1，我們棧是反著運算的，先計算1+1，然後計算1-2，結果就錯了，所以要先翻過來。

class Solution {
    public int calculate(String s) {
        Stack<Integer> numStack = new Stack<Integer>();
        Stack<Character> operaterStack = new Stack<Character>();
        
        char[]  chars = s.toCharArray();
        int i = 0;
        while(i<chars.length){
            if(chars[i]==' '){
                i++;
                continue;
            }
                
            else if(chars[i]=='+' || chars[i]=='-' || chars[i]=='*' || chars[i]=='/'){
                operaterStack.push(chars[i]);
                i++;
            }                
            else{
                int num = 0;
                while(i<chars.length && chars[i] != '+' && chars[i] !='-' && chars[i]!='*' && chars[i]!='/' && chars[i]!=' '){
                    num = num * 10 + chars[i] - '0';
                    i++;
                }
                if(operaterStack.empty() || operaterStack.peek()=='+' || operaterStack.peek()=='-')
                    numStack.push(num);
                else{
                    int num2 = numStack.pop();
                    char operator = operaterStack.pop();
                    if(operator == '*') numStack.push(num * num2);
                    else numStack.push(num2/num);
                }
            }
        }
        Stack<Integer> numStack2 = new Stack<Integer>();
        Stack<Character> operaterStack2 = new Stack<Character>();
        while(!numStack.empty()){
            numStack2.push(numStack.pop());
        }
        while(!operaterStack.empty()){
            operaterStack2.push(operaterStack.pop());
        }
        
        while(!operaterStack2.empty()){
            int num2 = numStack2.pop();
            int num1 = numStack2.pop();
            char operator = operaterStack2.pop();
            if(operator=='+') numStack2.push(num1 + num2);
            else numStack2.push(num2 - num1);
        }
        return numStack2.pop();
    }
}

228. Summary Ranges

這個題還是比較簡單的，用兩個指標，不過要注意的是當跳出迴圈的時候，還要再加上一個。

class Solution {
    public List<String> summaryRanges(int[] nums) {
        List<String> res = new ArrayList<String>();
        if(nums==null || nums.length==0)
            return res;
        int left = nums[0];
        int right = nums[0];
        for(int i=1;i<nums.length;i++){
            if(nums[i] - nums[i-1] == 1){
                right = nums[I];
            }
            else{
                if(left != right)
                    res.add(left + "->" + right);
                else
                    res.add(left+"");
                left = nums[I];
                right = nums[I];
            }
        }
        if(left != right)
            res.add(left + "->" + right);
        else
            res.add(left+"");
        return res;
    }
}

229. Majority Element II

Boyer-Moore演算法:比較直觀的解釋：在陣列中找到兩個不相同的元素並刪除它們，不斷重複此過程，直到陣列中元素都相同，那麼剩下的元素就是主要元素。
思想並不複雜，但是要憑空想出這個演算法來也不是件容易的事。另外，給我們的是陣列，直接在裡面刪除元素是很費時的。取而代之，可以利用一個計數變數來實現。

class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> list = new ArrayList();
        if(nums.length == 0) return list;
        int candidate1 = 0,candidate2 = 0;
        int count1 = 0, count2 = 0;
        
        for(int n : nums){
            if(candidate1 == n){
                count1++;
                continue;
            }
            else if(candidate2 == n){
                count2 ++;
                continue;
            }
            else if(count1 == 0){
                candidate1 = n;
                count1 = 1;
                continue;
            }
            else if(count2 == 0){
                candidate2 = n;
                count2 = 1;
                continue;
            }
            else{
                count1--;
                count2--;
            }
        }
        
        if (candidate1 == candidate2){
            list.add(candidate1);
            return list;
        }
        
        count1 = 0;
        count2 = 0;
        for (int n : nums) {
            if (n == candidate1)
                count1++;
            if (n == candidate2)
                count2++;
        }
        if (count1 > nums.length / 3)
            list.add(candidate1);
        if (count2 > nums.length / 3)
            list.add(candidate2);
        return list;
    }
}

230. Kth Smallest Element in a BST

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        if(root == null || k<=0)
            return -1;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode node = root;
        
        while(node!=null){
            stack.push(node);
            node = node.left;
        }
        
        int index = 0;
        TreeNode res = new TreeNode(-1);
        while(!stack.empty()){
            node = stack.pop();
            index++;
            System.out.println(index + "" + k);
            if(index==k){
                res = node;
                break;
            }
                
            node = node.right;
            while(node!=null){
                stack.push(node);
                node = node.left;
            }  
        }
        return res.val;
    }
}