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下面是今天写的几道题：
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.
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 }
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 -
这道题自己虽然知道要用位运算但是一开始没想出来，下边的方法借鉴网上，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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下边这个方法按道理是不符合题目的，题中要求不使用+-符号，但是这竟然在LeetCode上ac了，我这个代码需要小心边界即负数最小值。
1 //整数转换为二进制数组 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 //负数的话二进制取反+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 //负数需要反转+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 //两个二进制相加 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 //将二进制转换为十进制 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 //定义存储结果的sum，标识a，b的符号的falg 62 int sum = 0,flaga = 0,flagb = 0; 63 //定义存储ab以及相加后的二进制数组 64 int bita[33] = {0}; 65 int bitb[33] = {0}; 66 int sumbit[34] = {0}; 67 //考虑a，b可能为负数最小值 68 long long la = a; 69 long long lb = b; 70 //先按无符号位处理 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 //如果符号相同直接相加 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 //如果a是负数 94 else if(flaga == 1) 95 { 96 //先将a反转+1，此时a多了个符号位 97 reverdBitAdd1(bita,&sizea); 98 while(sizea<sizeb)//补齐负数位 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);//正数二进制位多 104 else if(sizea-1>sizeb)//负数符号位多 105 { 106 num--; 107 reverdBitAdd1(sumbit,&num); 108 sum = -sumBit(sumbit,num-1); 109 } 110 else if(sizea-1 == sizeb)//不包括符号位两个位数一样多 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 }
Nim Game，Reverse String，Sum of Two Integers
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原文地址：http://www.cnblogs.com/tenxi/p/5813853.html
