异或找唯一整数的程序是错的!
// https://stackoverflow.com/questions/41181004/bitwise-xor-operator-to-find-missing-unique-id public int findUniqueDeliveryId(int[] deliveryIds) { int uniqueDeliveryId = 0; for(int i = 0; i < deliveryIds.length; i++) { uniqueDeliveryId ^= deliveryIds[i]; } return uniqueDeliveryId; }
上面的程序是错的!
def findUniqueDeliveryId(deliveryIds): uniqueDeliveryId = 0; for i in deliveryIds: uniqueDeliveryId ^= i; # 我这里没错,我不用deliveryIds[i] for i in deliveryIds: print(i) return uniqueDeliveryId def test(l): print(l, findUniqueDeliveryId(l)) test([0, 1, 1]) # 0 test([3, 5, 5, 5]) # 6 test([3, 2, 2]) # 3
^, xor, exclusive or. 0^0=0, 0^1=1, 0^x=x, 1^0=1, 1^1=0.
异或没有进位这一说,和加法不同。把0^1^0^0^0这样的一串一位二进制数(二进制位)搞明白了,12 ^ 23 ^ 167这样的十进制数也就搞明白了——它们都是一串二进制位。每位都对则整个数对。
任意多个0异或之后还是0.
0 = 0
0 ^ 0 = 0
0 ^ 0 ^ 0 = 0
0 ^ 0 ^ 0 ^ 0 = 0
0 ^ 0 ^ 0 ^ 0 ^ 0 = 0
任意多个1异或,结果与1的个数有关。
1 = 1
1 ^ 1 = 0
1 ^ 1 ^ 1 = 1
1 ^ 1 ^ 1 ^ 1 = 0
1 ^ 1 ^ 1 ^ 1 ^ 1 = 1
111x11或00x00000按位异或,可分别看成x11111或x0000000按位异或。
1 ^ 偶数个1 = 1 ^ 0 = 1; 1 ^ 偶数个0 = 1 ^ 0 = 1; 别的情况基本都对。
1 ^ 偶数个1符合题意。3=011, 5=101,最低位都是1。6和7相同的位更多。和最低位(奇偶数)没关系。判断len是奇数偶数好像也没用。按题意,len>=3. 2个数谈不到unique.
def get_unique_num(ary) { if a[0] == a[1]: return a[2]; else # a[0] != a[1] #return a[0] if a[1] == a[2] else a[1] if a[1] == a[2]: return a[0]; # 1, 2, 2 else: # 2 1 2 return a[1] }
我这个也是错的,比如输入是2, 2, 2, 1, 打个补丁吧:
if a[0] == a[1]:
not_unique_num = a[0]
while a[i] == not_unique_num: ...
这个好像是对的。2, 2 .... 1,不管1有多远,反正2出现了至少两次,所以不是唯一的。un-unique, un-unravel :-)
用异或交换两个变量的值=奇技淫巧。嵌入式系统内存也没有那么紧张。STM32F103RBT6,主频 72MHz,128KB FLASH ,20KB RAM。1978年的Appe I是4KB RAM。[修改: 8051的基础内存为128字节]
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