257. Binary Tree Paths

类似于分治法吧。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res = new ArrayList<String>();
        if(root==null)
            return res;
        else{
            if(root.left == null && root.right == null)
                res.add(root.val+"");
            if(root.left!=null){
                List<String> left = binaryTreePaths(root.left);
                for(int i=0;i<left.size();i++){
                    res.add(root.val +"->" + left.get(i));
                }
            }

            if(root.right!=null){
                List<String> right = binaryTreePaths(root.right);
                for(int i=0;i<right.size();i++){
                    res.add(root.val +"->" + right.get(i));
                }
            }
        }
        return res; 
    }
}

258. Add Digits

小trick

class Solution {
    public int addDigits(int num) {
        int block = (num-1)/9;
        return num - (9*block);
    }
}

260. Single Number III

既然有两个数只出现了一次，那么这两个数至少有一位不一样，即异或不为0，那么可以根据这一位将数据分为两组，每一组再根据lian ge

class Solution {
    public int[] singleNumber(int[] nums) {
        int res = 0;
        for(int num:nums){
            res ^= num;
        }
        int seperator = 1;
        while((seperator & res) == 0){
            seperator <<= 1;
        }
        int[] result = new int[2];
        for(int num:nums){
            if((num & seperator) == 0)
                result[0] ^= num;
            else
                result[1] ^= num;
        }
        return result;
    }
}