LeetCode--Jewels and Stones && Range Sum of BST (Easy)

比眉伴天荒 2021-12-09 01:43 130阅读 0赞

# 771. Jewels and Stones (Easy) #

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;
    }

转载于:https://www.cnblogs.com/victorxiao/p/11093692.html