Bitwise AND of Numbers Range——LeetCode

class Solution {

public:

/*

mask的32位全为1，mask<<i是将mask的后i位置0，此时m或n和mask进行与操作，实际上就是将m或n的（副本）后i位置0。这时再进行异或操作，其实就是比较m和n的前32-i位是否相等，若相等，则停止循环，因为这相等的32-i位相与不会改变各个bit。又因m~n是连续变化的，后i位的各个bit为0和为1的情况都会出现，也就是说，m~n这些数相与后，后i位必定为0.综上，m和n后i位相与为0，前32-i位相同，从而m或n & mask<<i即是所求的值。

*/

int rangeBitwiseAnd(int m, int n) {

int i = 0, mask= ~0;

while ((m & mask<<i) ^ (n & mask<<i))

{

++i;

}

return m & mask<<i ;

}

};

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时间： 2024-08-03 20:59:34

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