求最小的两个数相加为sum

//求最小的两个数相加为sum

    public ArrayList<Integer> FindNumbersWithSum(int [] array,int sum) {

       List<Integer> list = new ArrayList<Integer>();

        if(array == null || array.length==0){

            return (ArrayList<Integer>) list;

        }

        for(int i=0;i<array.length-1;i++){

            for(int j=i+1;j<array.length;j++){

                if(array[i]+array[j] == sum){

                   list.add(array[i]);

                    list.add(array[j]);

                    break;

                }

            }

            if(list.size()!=0){

                break;

            }

        }

        //没找到就返回空集合

        return (ArrayList<Integer>) list;

    }

计数一个int型的二进制有多少个1，难点在负数需要补码存

    public int MoreThan0(int n) {

        int cnt = 0;

        char []c = Integer.toBinaryString(n).toCharArray();

        for (char d : c) {

            if(d=='1'){

                cnt++;

            }

        }

        return cnt;

    }

求一个二叉树是否左右对称

中心处理方法
- 思路，利用栈队列的特性，进行层比较



import java.util.LinkedList;

import java.util.Stack;

public class Solution {

    boolean isSymmetrical(TreeNode pRoot) {

        Stack<TreeNode> stack = new Stack<TreeNode>();

        Stack<TreeNode> StackSon = new Stack<TreeNode>();

        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();

        stack.push(pRoot);

        queue.offer(pRoot);

        TreeNode bt1 = null;

        TreeNode bt2 = null;

        while(stack.size()!=0 || queue.size()!=0){

            while(stack.size()!=0){

                //出队和出栈

                bt1 = stack.pop();

                bt2 = queue.poll();

                //比较镜像元素

                if(bt1==null || bt2==null){

                    break;

                }

                if(bt1.val != bt2.val){

                    return false;

                }

                //入队

                queue.offer(bt1.right);

                queue.offer(bt1.left);

                //栈的中转队列

                StackSon.push(bt2.right);

                StackSon.push(bt2.left);

            }

            while(StackSon.size()!=0){

                stack.push(StackSon.pop());

            }

        }

        return true;

    }

    public static void main(String[] args) {

        int a[] = {8,6,6,5,7,7,5};

        TreeNode root= new CreatTree().CreatTreeWay(a);

        System.out.println(new Solution().isSymmetrical(root));

    }

}

按层创建二叉树



import java.util.LinkedList;

public class CreatTree {

    TreeNode root = null;

    public TreeNode CreatTreeWay(int a[]){

        if(a==null || a.length==0 ){

            return null;

        }

        if(root == null){

            root = new TreeNode(a[0]);

        }else{

            return root;

        }

        LinkedList<TreeNode> list = new LinkedList<TreeNode>();

        list.offer(root);

        TreeNode node = null;

        for (int i = 1; i < a.length && list.size()!=0; i++) {

            node = list.poll();

            node.left = new TreeNode(a[i]);

            list.offer(node.left);

            if(i+1<a.length){

                node.right = new TreeNode(a[i+1]);

                list.offer(node.right);

            }

        }

        return root;

    }

    public static void main(String[] args) {

        int a[] = {8,6,6,5,7,7,5};

        new CreatTree().CreatTreeWay(a);

    }

}

忘了提供二叉树的结点类了



public class TreeNode {

    int val = 0;

    TreeNode left = null;

    TreeNode right = null;

    public TreeNode(int val) {

        this.val = val;

    }

}

巴特西

牛客网2016.4.11（两个数相加为sum/计数一个int型的二进制有多少个1/二叉树是否左右对称）

求最小的两个数相加为sum

计数一个int型的二进制有多少个1，难点在负数需要补码存

求一个二叉树是否左右对称

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