leetcode水题题解

344. Reverse String

Write a function that takes a string as input and returns the string reversed.

Example:
Given s = "hello", return "olleh".

class Solution {
public:
	string reverseString(string s) {
		string::size_type i, j;
		if (s.size() == 0)
		{
			return s;
		}
		for (i = 0, j = s.size() - 1; i < j; i++, j--)
		{
			char ch = s[i];
			s[i] = s[j];
			s[j] = ch;
		}
		return s;
	}
};

记住size_type是unsigned的就可以了

412. Fizz Buzz

Write a program that outputs the string representation of numbers from 1 to n.

But for multiples of three it should output “Fizz” instead of the number and for the multiples of five output “Buzz”. For numbers which are multiples of both three and five output “FizzBuzz”.

Example:

n = 15,

Return:
[
    "1",
    "2",
    "Fizz",
    "4",
    "Buzz",
    "Fizz",
    "7",
    "8",
    "Fizz",
    "Buzz",
    "11",
    "Fizz",
    "13",
    "14",
    "FizzBuzz"
]

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> vec;
        for (int i = 1; i <= n; ++i)
        {
        	if (i % 3 == 0 && i % 5 == 0)
        	{
        		vec.push_back("FizzBuzz");
        	}
        	else if (i % 3 == 0)
        	{
        		vec.push_back("Fizz");
        	}
        	else if(i % 5 == 0)
        	{
        		vec.push_back("Buzz");
        	}
        	else
        	{
        		stringstream ss;
        		ss << i;
        		vec.push_back(ss.str());
        	}
        }
        return vec;
    }
};

292. Nim Game

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.


Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.


For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.


Hint:

If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner?

最开始的想法：

保存先前的结果，也就是记忆化。

1:win
2:win
3:win
4:lost
5:win
6:win
7:win
8:lost
9:win
10:win
11:win
12:lost
13:win
14:win
15:win
16:lost
17:win
18:win
19:win
20:lost
21:win
22:win
23:win
24:lost
25:win
26:win
27:win
28:lost
29:win
30:win
31:win
32:lost
[Finished in 1.5s]

2 3 4一组，3 4 5一组， 4 5 6一组，这样。

n减去每一组，和先前的结果建立联系。

#include <iostream>
#include <vector>
#include <set>
using namespace std;

class Solution {
public:
	Solution()
	{
		for (int j = 7; j <= 100; ++j)
		{
			canWinNim(j);
		}
	}
	bool canWinNim(int n) {
		int i;
		bool flag;
		
		s.insert(1);
		s.insert(2);
		s.insert(3);
		s.insert(5);
		s.insert(6);
		//1 2 3 4 5 6特殊处理
		if ((n >= 1 && n <= 3) || n == 5 || n == 6)
		{
			return true;
		}
		if (n == 4)
		{
			return false;
		}
		//1
		flag = true;
		for (i = 2; i <= 4; ++i)
		{
			if (s.count(n - i) == 0)
			{
				flag = false;
				break;
			}
		}
		if (flag)
		{
			s.insert(n);
			return true;
		}
		//2
		flag = true;
		for (i = 3; i <= 5; ++i)
		{
			if (s.count(n - i) == 0)
			{
				flag = false;
				break;
			}
		}
		if (flag)
		{
			s.insert(n);
			return true;
		}
		//3
		flag = true;
		for (i = 4; i <= 6; ++i)
		{
			if (s.count(n - i) == 0)
			{
				flag = false;
				break;
			}
		}
		if (flag)
		{
			s.insert(n);
			return true;
		}
		return false;
	}
private:
	set<int>s;
};
int main()
{
	Solution s;
	for (int i = 1; i <= 32; ++i)
	{
		if (s.canWinNim(i))
		{
			cout << i << ":win" << endl;
		}
		else
		{
			cout << i << ":lost" << endl;
		}
	}
	//int x = 1199886170;
			//2147483647
	return 0;
}

超时，在这个点1199886170

必然，预处理得超过它。

思考下：但凡两人同时取过过后，剩下的是4，先取的就输了

由上图，对于8，先取的输了

由上图，对于12，先取的输了

16， 20， 24，必然都是先取的输了

class Solution {
public:
	
	bool canWinNim(int n) {
        if(n % 4 == 0)
        {
            return false;
        }
        return true;
	}

};

367. Valid Perfect Square

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16
Returns: True

Example 2:

Input: 14
Returns: False

二分查找，对于整数相乘溢出，得做个检测

。

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <string>
#include <sstream>
#include <cstring>
using namespace std;

class Solution {
public:
    bool isPerfectSquare(int x) {
        if (x == 0 || x == 1)
        {
            return true;
        }
        return BinarySearch(x, x);
    }
private:
    bool BinarySearch(int x, int target)
    {
        int left = 1;
        int right = x / 2;
        while(left <= right)
        {
            int mid = (left + right) / 2;
            //溢出
            if (mid * mid < 0)
            {
                right = mid - 1;
            }
            else if (mid * mid > target || target / mid < mid)
            {
                right = mid - 1;
            }
            else if(mid * mid < target)
            {
                left = mid + 1;
            }
            else if(mid * mid == target)
            {
                return true;
            }
        }
        return false;
    }
};

int main()
{
	Solution s;
    cout << s.isPerfectSquare(24);
	return 0;
}