LeetCode Weekly Contest 19

LeetCode Weekly Contest 19

504. Base 7

My SubmissionsBack To Contest

 

• User Accepted: 994
• User Tried: 1054
• Total Accepted: 1345
• Total Submissions: 3160
• Difficulty: Easy

Given an integer, return its base 7 string representation.

Example 1:

Input: 100
Output: "202"

Example 2:

Input: -7
Output: "-10"

Note: The input will be in range of [-1e7, 1e7].

简单的十进制转7进制。

class Solution {
public:
    string convertToBase7(int num) {
        if(num == 0){
            return "0";
        }
        string s, p=""; 
        if(num < 0){
            p = "-"; 
            num  = -num; 
        }
        while(num){
            s = char('0' + (num%7))  + s; 
            num = num/7; 
        }
        return p+s; 
    }
};

　　

513. Find Bottom Left Tree Value

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• User Accepted: 870
• User Tried: 943
• Total Accepted: 1060
• Total Submissions: 2399
• Difficulty: Medium

Given a binary tree, find the leftmost value in the last row of the tree.

Example 1:

Input:

    2
   / 
  1   3

Output:
1

Example 2: 

Input:

        1
       / 
      2   3
     /   / 
    4   5   6
       /
      7

Output:
7

Note: You may assume the tree (i.e., the given root node) is not NULL.

树结构的迭代

/**
 * 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 ans, current; 
    void Find(TreeNode *root, int row){
            if(row >= current){
                ans = root->val; 
                current = row; 
            }
        if(root->right != NULL){
            Find(root->right, row+1); 
        }
        if(root->left != NULL){
            Find(root->left, row+1); 
        }
    }
    
    int findBottomLeftValue(TreeNode* root) {
        ans = root->val; current = 1; 
        Find(root, 1); 
        return ans; 
    }
};

　　

515. Find Largest Value in Each Tree Row

My SubmissionsBack To Contest

 

• User Accepted: 851
• User Tried: 878
• Total Accepted: 943
• Total Submissions: 2140
• Difficulty: Medium

You need to find the largest value in each row of a binary tree.

Example:

Input: 

          1
         / 
        3   2
       /      
      5   3   9 

Output: [1, 3, 9]

也是用树结构，使用 BFS

/**
 * 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:
    vector<int> largestValues(TreeNode* root) { 
        vector<int> ans; 
        if(root == NULL){
            return ans; 
        }
        
        TreeNode *tmp; 
        queue<TreeNode* > q; 
        queue<int> qlevel; 
        
        qlevel.push(1); 
        q.push(root); 
        
        int cur_level = 1, cur_val = -2147483648; 
        while(!q.empty()){
            tmp = q.front(); q.pop(); 
            int tmp_level = qlevel.front(); qlevel.pop(); 
            
            if(tmp_level == cur_level){
                if(cur_val < tmp->val){
                    cur_val = tmp->val; 
                }
            }else{
                ans.push_back( cur_val ); 
                cur_level = tmp_level; 
                cur_val = tmp->val; 
            }
            
            if(tmp->left != NULL){
                q.push(tmp->left); qlevel.push(tmp_level + 1); 
            }
            if(tmp->right != NULL){
                q.push(tmp->right); qlevel.push(tmp_level + 1); 
            }
        }
        ans.push_back(cur_val); 
        
        return ans; 
    }
};

　　

493. Reverse Pairs

My SubmissionsBack To Contest

 

• User Accepted: 165
• User Tried: 777
• Total Accepted: 172
• Total Submissions: 3595
• Difficulty: Hard

Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j].

You need to return the number of important reverse pairs in the given array.

Example1:

Input: [1,3,2,3,1]
Output: 2

Example2:

Input: [2,4,3,5,1]
Output: 3

Note:

1. The length of the given array will not exceed 50,000.
2. All the numbers in the input array are in the range of 32-bit integer.

看了discuss 才有思路， 原来 merge sort 也有这么好功能

class Solution {
private:
    int cnt; 
    
    void countPair(vector<int>& nums, int lf, int md, int rg){
        int l = lf, r = md + 1; 
        while(l <= md && r <= rg){
            if((long)nums[l] > (long)2*nums[r]){
                cnt += (md - l + 1); 
                r++; 
            }else{
                l++; 
            }
        }
        sort(nums.begin() + lf, nums.begin() + rg + 1); 
    }
    
    void Reverse(vector<int>& nums, int lf, int rg){
        if(lf == rg){
            return; 
        }
        int mid = lf + (rg - lf)/2; 
        Reverse(nums, lf, mid); 
        Reverse(nums, mid+1, rg); 
        
        countPair(nums, lf, mid, rg); 
    }
    
    
    
public:
    int reversePairs(vector<int>& nums) {
        int len = nums.size(); 
        if(len <= 1){
            return 0; 
        }
        cnt = 0; 
        Reverse(nums, 0, len-1); 
        return cnt; 
    }
};