对于初学者,我确实看过这些问题:
Algorithm to find two repeated numbers in an array, without sorting
这个不同:
给出一个带有一个唯一编号的未排序整数数组,其余数字重复3次, 即:
{4,5,3, 5,3,4, 1, 4,3,5 }
我们需要在O(n)时间和O(1)空间中找到这个唯一的数字
注意:这不是作业,只是我遇到一个很好的问题
答案 0 :(得分:9)
这个怎么样:
想法:按位添加mod 3
#include <stdio.h>
int main() {
int a[] = { 1, 9, 9, 556, 556, 9, 556, 87878, 87878, 87878 };
int n = sizeof(a) / sizeof(int);
int low = 0, up = 0;
for(int i = 0; i < n; i++) {
int x = ~(up & a[i]);
up &= x;
x &= a[i];
up |= (x & low);
low ^= x;
}
printf("single no: %d\n", low);
}
答案 1 :(得分:0)
此解决方案适用于所有输入。 这个想法是从数组中提取一个整数的位并加到各自的32bit 位图“ b”(实现为32字节数组,表示32位编号)
unsigned int a[7] = {5,5,4,10,4,9,9};
unsigned int b[32] = {0}; //Start with zeros for a 32bit no.
main1() {
int i, j;
unsigned int bit, sum =0 ;
for (i=0;i<7; i++) {
for (j=0; j<32; j++) { //This loop can be optimized!!!!
bit = ((a[i] & (0x01<<j))>>j); //extract the bit and move to right place
b[j] += bit; //add to the bitmap array
}
}
for (j=0; j<32; j++) {
b[j] %= 2; //No. repeating exactly 2 times.
if (b[j] == 1) {
sum += (unsigned int) pow(2, j); //sum all the digits left as 1 to get no
//printf("no. is %d", sum);
}
}
printf("no. is %d", sum);
}