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;
}