Nim Game，Reverse String，Sum of Two Integers

 下面是今天寫的幾道題：

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.

1 bool canWinNim(int n) {
2   if ( 0<n && n<4 ) return true;
3   else if(n==4) return false;
4   else return n%4 != 0;
5 }

 

344. Reverse String

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

1.C++ code  Runtime：12ms

1 class Solution {
2 public:
3     string reverseString(string s) {
4        reverse(s.begin(),s.end());
5        return s;
6     }
7 };

2.C code  Runtime：4ms

 1 char* reverseString(char* s) {
 2     int len = strlen(s);
 3     for(int i=0;i<len/2;i++)
 4     {
 5         char tempc = s[i];
 6         s[i] = s[len-1-i];
 7         s[len-1-i] = tempc;
 8     }
 9     return s;
10 }

371. Sum of Two Integers

Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -

這道題本身雖然知道要用位運算可是一開始沒想出來，下邊的方法借鑑網上，原文地址http://www.cnblogs.com/la0bei/p/5659829.html

 1 int main()
 2 {
 3        int a,b,c;
 4        cin >> a >> b;
 5        while(b)
 6        {
 7               c = a^b;
 8               b = (a&b)<<1;
 9               a = c;
10        }
11        cout << a << endl;
12        system("pause");
13 }

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下邊這個方法按道理是不符合題目的，題中要求不使用+-符號，可是這居然在LeetCode上ac了，我這個代碼須要當心邊界即負數最小值。

  1 //整數轉換爲二進制數組
  2 int intToBit(int *str,long long num)
  3 {
  4     int i=0;
  5     do
  6     {
  7         str[i++] = num%2;
  8         num = num/2;
  9     }while(num>0);
 10 
 11     return i;
 12 }
 13 //負數的話二進制取反+1
 14 void reverdBitAdd1(int *str,int *num)
 15 {
 16     //取反
 17     int temp = *num;
 18     while(temp--) str[temp] = ( (str[temp] == 0)? 1 : 0 );
 19 
 20     //負數須要反轉+1;
 21     int sum = 1;
 22     for(int i=0;i<*num;i++)
 23     {
 24         sum += str[i];
 25         str[i] = sum%2;
 26         sum = sum/2;
 27     }
 28     str[*num] = 1;
 29     (*num)++;
 30 }
 31 
 32 //兩個二進制相加
 33 int bitAdd(int *sumbit,int*a,int sizea,int*b,int sizeb)
 34 {
 35     int sum = 0;
 36     int num = (sizea>sizeb?sizea:sizeb);
 37     for (int i=0;i<num;i++)
 38     {
 39         sum +=(a[i]+b[i]);
 40         sumbit[i] = sum%2;
 41         sum /=2;
 42     }
 43     if (sum ==1) sumbit[num++] = 1;
 44     return num;
 45 }
 46 
 47 //將二進制轉換爲十進制
 48 int sumBit(int *bitNum, int n)
 49 {
 50     int sum = 0;
 51     int bit2 = 1;
 52     for(int i=0;i<n;i++)
 53     {
 54         sum += bitNum[i]*bit2;
 55         bit2 *=2;
 56     }
 57     return sum;
 58 }
 59 
 60 int getSum(int a, int b) {
 61     //定義存儲結果的sum，標識a，b的符號的falg
 62     int sum = 0,flaga = 0,flagb = 0;
 63     //定義存儲ab以及相加後的二進制數組
 64     int bita[33] = {0};
 65     int bitb[33] = {0};
 66     int sumbit[34] = {0};
 67     //考慮a，b可能爲負數最小值
 68     long long la = a;
 69     long long lb = b;
 70     //先按無符號位處理
 71     if(la<0)
 72     {
 73         la = -la;
 74         flaga = 1;
 75     }
 76 
 77     if(lb<0)
 78     {
 79         lb = -lb;
 80         flagb = 1;
 81     }
 82 
 83     int sizea = intToBit(bita,la);
 84     int sizeb = intToBit(bitb,lb);
 85 
 86     //若是符號相同直接相加
 87     if(flaga == flagb)
 88     {
 89         int num = bitAdd(sumbit,bita,sizea,bitb,sizeb);
 90         sum = sumBit(sumbit,num);
 91         if(flaga == 1) sum = -sum;
 92     }
 93     //若是a是負數
 94     else if(flaga == 1)
 95     {
 96         //先將a反轉+1，此時a多了個符號位
 97         reverdBitAdd1(bita,&sizea);
 98         while(sizea<sizeb)//補齊負數位
 99         {
100             bita[sizea++] = 1;
101         }
102         int num = bitAdd(sumbit,bita,sizea,bitb,sizeb);
103         if(sizeb>sizea-1) sum = sumBit(sumbit,num-1);//正數二進制位多
104         else if(sizea-1>sizeb)//負數符號位多
105         {
106             num--;
107             reverdBitAdd1(sumbit,&num);
108             sum = -sumBit(sumbit,num-1);
109         }
110         else if(sizea-1 == sizeb)//不包括符號位兩個位數同樣多
111         {
112             if(sumbit[num] == 1) sum = -sumBit(sumbit,num-1);
113             else sum = sumBit(sumbit,num-1);
114         }
115     }
116     else if(flagb == 1)
117     {
118         reverdBitAdd1(bitb,&sizeb);
119         while(sizeb<sizea)
120         {
121             bitb[sizeb++] = 1;
122         }
123         int num = bitAdd(sumbit,bita,sizea,bitb,sizeb);
124         if(sizea>sizeb-1) sum = sumBit(sumbit,num-1);
125         else if(sizeb-1>sizea)
126         {
127             num--;
128             reverdBitAdd1(sumbit,&num);
129             sum = -sumBit(sumbit,num-1);
130         }
131         else if(sizeb-1 == sizea)
132         {
133             if(sumbit[num] == 1) sum = -sumBit(sumbit,num-1);
134             else sum = sumBit(sumbit,num-1);
135         }
136     }
137     return sum;
138 }