两道笔试题

Write an algorithm to determine if a number is "happy".
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
Example: 19 is a happy number
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1


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public class Solution {
     
    private static int maxTimes = 100;
         
    public boolean isHappy(int n, int times) {
        String sn = String.valueOf(n);
        int ss = sn.length();
        int sum = 0;
        if(times >= Solution.maxTimes){
           return false;
        }
        for(int i=0;i<ss-1; i++){
           int y = n % Math.pow(10, ss-i);
           int z = y / Math.pow(10, i+1);
           sum += Math.pow(z, 2);
        }
        if(sum == 1) return true;
        times++;
        isHappy(sum, times);
    }
}

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Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
Example 1:
    2
   / \
  1   3
Binary tree [2,1,3], return true.
Example 2:
    1
   / \
  2   3
Binary tree [1,2,3], return false.


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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null) return false;
        if(root.left != null){
           if(root.left.val >= root.val)return false;
        }
        if(root.left.right != null){
           if(root.left.right.val >= root.val)return false;
        }
        if(root.right != null){
           if(root.right.val <= root.val)return false;
        }
        if(root.right.left != null){
           if(root.right.left.val <= root.val)return false;
        }
        if(root.left != null) isValidBST(root.left);
        if(root.right != null) isValidBST(root.right);
        return true;
    }
}