970.Powerful Integers - Leetcode题解

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9.Palindrome Number

13.Roman to Integer

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20.Valid Parentheses

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219.Contains Duplicate II

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226.Invert Binary Tree

231.Power of Two

232.Implement Queue Using Stack

234.Palindrome Linked List

237.Delete Node in a Linked List

240.Search a 2D Matrix II

242.Valid Anagram

257.Binary Tree Paths

263.Ugly Numbers

268.Missing Number

278.First Bad Version

284.Peeking Iterator

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344.Reverse String

345.Reverse Vowels of a String

349.Intersection of Two Arrays

350.Intersection of Two Arrays II

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429.N Ary Tree Level Order Traversal

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441.Arranging Coins

443.String Compression

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669.Trim a Binary Search Tree

671.Second Minimum Node in a Binary Tree

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687.Longest Univalue Path

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720.Longest Word in Dictionary

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766.Toeplitz Matrix

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970.Powerful Integers

974.Subarray Sums Divisible By K

976.Largest Perimeter Triangle

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1008.Construct Binary Search Tree from Preorder Traversal

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面试题 16.01

面试题 16.18

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970.Powerful Integers

970.Powerful Integers
难度:Easy
Given two positive integers x and y, an integer is powerful if it is equal to x^i + y^j for some integers i >= 0 and j >= 0.
Return a list of all powerful integers that have value less than or equal to bound.
You may return the answer in any order. In your answer, each value should occur at most once.
Example 1:
​
Input: x = 2, y = 3, bound = 10
Output: [2,3,4,5,7,9,10]
Explanation: 
2 = 2^0 + 3^0
3 = 2^1 + 3^0
4 = 2^0 + 3^1
5 = 2^1 + 3^1
7 = 2^2 + 3^1
9 = 2^3 + 3^0
10 = 2^0 + 3^2
Example 2:
​
Input: x = 3, y = 5, bound = 15
Output: [2,4,6,8,10,14]
 
​
Note:
​
1 <= x <= 100
1 <= y <= 100
0 <= bound <= 10^6
直观暴力解法，需要考虑特殊情况，如x、y为1的情况。
class Solution {
vector<int> in={2};
bool isValid(int x,int y,int num)
{
    if(x==1 && y==1) return num==2;
    if(num==x+1 || num == y+1) return true;
    int lenx=log(double(num))/log(double(x));
  //  cout <<lenx<<endl;
    for(int i=0;i<=lenx;i++)
    {
        int tmp=num-pow(x,i);
        if(tmp==0) return false;
       if( pow(y,int(log(double(tmp))/log(double(y)))) ==tmp)  return true;
    }
    return false;
}
public:
    vector<int> powerfulIntegers(int x, int y, int bound) {
        vector<int>res;
        if(bound<2) return res;
        if(x==1&& y==1) 
        {
           res.push_back(2);
            return res;
        }
        if(x<y) swap(x,y);
        if(y==1) 
        {
            bound-=1;
            int lenx= log(double(bound))/log(double(x)) ;
            for(int i=0;i<=lenx;i++)
                res.push_back(pow(x,i)+1);
            return res;
        }
        res.push_back(2);
        for(int start=y+1;start<=bound;start++)
            if(isValid(x,y,start))
                res.push_back(start);
       // cout<<isValid(x,y,bound)<<endl;
        return res;
    }
};
执行用时 :1084 ms, 在所有 C++ 提交中击败了5.60%的用户
内存消耗 :8.6 MB, 在所有 C++ 提交中击败了80.70%的用户
两重遍历，长度是对数。另外需要使用集合去重。
class Solution {
public:
    vector<int> powerfulIntegers(int x, int y, int bound) {
        if(bound<2) return vector<int>();
        int xm= x==1 ? 1: log(double(bound))/log(double(x));
        int ym = y==1? 1: log(double(bound))/log(double(y));
       // cout<<xm<<" "<<ym<<endl;
        unordered_set<int>all;
        for(int i=0;i<=xm;i++)
        {
        for(int j=0;j<=ym;j++)
        {
            int tmp=pow(x,i)+pow(y,j);
      //     cout<<tmp<<endl;         
            if(tmp>bound) 
                break;     
            else
                if(!all.count(tmp))all.insert(tmp);
        }            
        }
        return vector<int>(all.begin(),all.end());
    }
};
执行用时 :4 ms, 在所有 C++ 提交中击败了91.60%的用户
内存消耗 :8.8 MB, 在所有 C++ 提交中击败了78.07%的用户

965.Univalued Binary Tree

974.Subarray Sums Divisible By K

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