You're given strings J representing the types of stones that are jewels, and S representing the stones you have.  Each character in S is a type of stone you have.  You want to know how many of the stones you have are also jewels.

The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so "a" is considered a different type of stone from "A".

Example 1:

Input: J = "aA", S = "aAAbbbb"
Output: 3

Example 2:

Input: J = "z", S = "ZZ"
Output: 0

Note:

S and J will consist of letters and have length at most 50.
The characters in J are distinct.

solution##

class Solution {
    public int numJewelsInStones(String J, String S) {
    int count=0;
    for(int i=0; i<J.length(); i++)
    {
        for(int j=0; j<S.length(); j++)
        {
            if(J.charAt(i) == S.charAt(j))
                count++;
        }
     }
     return count;
   }
}

总结##

此题思路较简单，用两个循环加一个计数器count解决即可，外层循环遍历J，内层循环遍历S。先取J里面的一个字符与S里面的字符挨个比较，如果两字符相同，计数器count++，否则什么都不做，这样直至循环结束。

938. Range Sum of BST（Easy）#

Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).

The binary search tree is guaranteed to have unique values.

 
Example 1:

Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
Output: 32
Example 2:

Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
Output: 23
 
Note:

The number of nodes in the tree is at most 10000.
The final answer is guaranteed to be less than 2^31.

solution##

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int rangeSumBST(TreeNode root, int L, int R) {
        int leftsum = 0; int rightsum = 0;
        if (root.left != null)
            leftsum = rangeSumBST(root.left,L,R);
        if (root.right != null)
            rightsum = rangeSumBST(root.right,L,R);
        if (L <= root.val && root.val <= R)
            return root.val + leftsum + rightsum;
        return leftsum + rightsum;
    }
}

总结##

此题为二叉树结点值求和的升级版。整体思想为先求左子树结点值大于L小于R的和，再求右子树结点值大于L小于R的和，最后，如果根结点值大于L小于R，则加上根结点值。
首先定义两个int变量leftsum和rightsum用来分别存放左右子树的和，如果左子树非空，则递归求左子树结点值的和leftsum，如果右子树非空，则递归求右子树结点值的和rightsum。最后，如果根结点值大于L小于R，则加上根结点值并返回，否则，只返回leftsum+rightsum的值。

求二叉树结点里面值（假设为int）的和

public int sumBST(TreeNode root) 
{
    int leftsum = 0; int rightsum = 0;
    if (root.left != null)
        leftsum = sumBST(root.left);
    if (root.right != null)
        rightsum = sumBST(root.right);
    return root.val + leftsum + rightsum;
}