每日一题计划

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LeetCode难度三个等级：

1，简单：Easy

1，中等：Medium

3，困难：Easy

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2019-06-10

题目链接 771.Jewels and Stones

Easy

水题，就是求字符串\(S\)中有多少字母在字符串\(J\)中，字符串\(J\)中的字母各不相同。

代码如下:复杂度\(O(S.length+J.length)\)。

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        int sum=0;
        unordered_set<char>se(J.begin(),J.end());
        for(auto i : S) sum+=se.count(i);
        return sum;
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Jewels and Stones.

Memory Usage: 8.6 MB, less than 56.84% of C++ online submissions for Jewels and Stones.

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2019-06-11

题目链接938.Range Sum of BST

Easy

题意就是题目的名字，给你一个建好的二叉搜索树(\(BST\))的根节点\(root\)，让你计算值位于\(L\)和\(R\)之间节点的值的和。
\[ ans=\sum\{node.value\ which\ L\le node.value\le R\} \]
然后代码就是一个简单的后序遍历加类似于二分的剪枝。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int rangeSumBST(TreeNode* root, int L, int R) {
        int sum=0;
        if(root){
            if(root->val>=L) sum+=rangeSumBST(root->left,L,R);
            if(root->val<=R) {
                sum+=rangeSumBST(root->right,L,R);
                if(root->val>=L) sum+=root->val;
            }
        }
        return sum;
    }
};

Runtime: 144 ms, faster than 90.39% of C++ online submissions for Range Sum of BST.

Memory Usage: 41.3 MB, less than 41.58% of C++ online submissions for Range Sum of BST.