LeetCode 201. Bitwise AND of Numbers Range

可通过的代码：

class Solution {public:int rangeBitwiseAnd(int m, int n){int ret = 0;for (int i = 0; m!=0 && n!=0 && i<31; n>>=1, m>>=1, i++){ret += ((m%2!=0)&&m==n? (1<<i): 0);}return ret;}};在VS2013下测试成功(n=1, m=1, 输出1)，提交到网上测试失败(n=1, m=1, 提示输出0)的代码：class Solution {public:int rangeBitwiseAnd(int m, int n){int ret = 0;for (int i = 0; i < 31; ++ i){ret |= (((n-m)>(1<<(i+1)) || (n/(1<<i))%2==0 || (m/(1<<i))%2==0)? 0: (1<<i));}return 0;}};

，这一秒不放弃，下一秒就会有希望。