[LeetCode] Weekly Contest 150 – Maximum Level Sum of a Binary Tree

發佈日期: 作者: user

Leetcode又荒廢了好久XD

之後應該會多寫DP的題目，看同學在US刷題蠻多在看DP的題目

https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree

找出BT裡面最大SUM的ROW，這題應該算簡單的吧，才給四分。

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class Solution {
public:
    int maxLevelSum(TreeNode* root) {
        
        int row_sum = 0, ans_sum = 0, level = 0, ans_level = 0, ans = 0;
        vector<vector<TreeNode*>> vec;
        vec.push_back(vector<TreeNode*>());
        vec.push_back(vector<TreeNode*>());
        
        if(root != NULL)
            vec[level].push_back(root);
        else
            return 0;
 
        while (!vec[(level+1)%2].empty() || !vec[level%2].empty()) {
            while (!vec[level%2].empty()) {
                int len = vec[level%2].size();
            
                for (int i = 0; i < len; i++)
                {
                    row_sum += vec[level%2][i]->val;
 
                    if (vec[level%2][i]->left != NULL)
                        vec[(level+1)%2].push_back(vec[level%2][i]->left);
                    
                    if (vec[level%2][i]->right != NULL)
                        vec[(level+1)%2].push_back(vec[level%2][i]->right);
                }
 
                if (row_sum > ans_sum) {
                    ans_sum = row_sum;
                    ans = level+1;
                }
                
                vec[level%2].clear();
                level++;
                row_sum = 0;
            }
        }
        return ans;
        
    }
};

用兩個queue交互儲存下一個row的nodes，並計算當前row的sum，看有沒有大於temp ans

大概就是這樣，但這版Code的可讀性應該蠻差的…，看看有沒有機會再改。

[LEETCODE] Longest Substring Without Repeating Characters

發佈日期: 作者: user

https://leetcode.com/problems/longest-substring-without-repeating-characters/

第一版

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class Solution {
public:
    string cs;
    
    int lengthOfLongestSubstring(string s) {
        int l = s.length();
        int ans = 0;
        
        cs = s;
        
        for (int i = 0; i < l; i++) {
           ans = max(ans, checkNonDoublealphabets(i-ans,i));
        }      
        
        return ans;
    }
    
    int checkNonDoublealphabets(int s, int e)
    {        
        bool alphabets[256] = {false};        
 
        
        for (int i = s; i <= e; i++) {
            if(alphabets[static_cast<int>(cs[i])] == true)
                return 0;
            alphabets[static_cast<int>(cs[i])] = true;
        }
        return e-s+1;
    }
};

先做了一版slide window, length = ans, 每次都往前檢查 window內是不是有重複的字元

有的話就退出，沒有的話就把新的字元吃進來=>長度加一

大概只有50%的ranking…

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