Q5.7 Find the missing number

Q: An array A[1…n] contains all the integers from 0 to n except for one number which is missing. In this problem, we cannot access an entire integer in A with a single operation. The elements of A are represented in binary, and the only operation we can use to access them is “fetch the jth bit of A[i]”, which takes constant time. Write code to find the missing integer. Can you do it in O(n) time?

A:

思路1：通过fetch将A[i]的每一位都可以取出，那么就很容易得到A[i]的值，有了值之后将数组求和，其与n*(n+1)/2 的差值就是丢失的数字。

思路2：思路1的方法很直观，但是效率还不够高。

以n = 13为例：

00000          00100          01000          01100

00001          00101          01001          01101

00010          00110          01010 

---------          00111          01011

可以发现其最低有效位LSB1 有则一定的规律

if n%2 == 1 then count(0s) = count(1s)

if n%2 == 0 then count(0s) = 1 + count(1s)

得到这样一个表格：

	count(0s) = 1 + count(1s)	count(0s) =  count(1s)
LSB(v) == 0	所以少了一个0，则count(0s) = counti(1s)	所以少了一个0，则count(0s) < ounti(1s)
LSB(v) == 1	所以少了一个1，则count(0s) > counti(1s)	所以少了一个0，则count(0s) > counti(1s)

if LSB(v) == 0, then 去除LSB为1的数据，对余下的数据对下一位的bit位进行判断

if LSB(v) == 1, then 去除LSB为0的数据，对余下的数据对下一位的bit位进行判断.

还是以例子说明，其最后一位count(0s) > count(1s), 由表格知道LSB(v)为1，则去除LSB为0的数据，即如下：

00000          00100          01000          01100

00001          00101          01001          01101

00010          00110          01010 

---------          00111          01011

对余下的数据统计其倒数第二位count(0s)和count(1s), 比较其大小，得到丢失数v的倒数第二位，并进一步删除数据。这样递归下去。

#include <iostream>
#include <vector>
using namespace std;

int fetch(int n, int i) {
	return (n>>i)&1;
}

int getNum(int n) {
	int ret = 0;
	for (int i = 31; i >= 0; i--) {
		ret = (ret<<1)|fetch(n,i);
	}
	return ret;
}

int findMissing1(int n, int a[]) {
	int sum = 0;
	int len = 8*sizeof(int);
	for (int i = n-1; i >= 0; i--) {
		sum += getNum(a[i]);
	}
	for (int i = 1; i <= n; i++) {
		sum -= i;
	}
	return -sum;	
}

int findMissing2(vector<int> &array, int ind) {
	if (ind >= 32) {
		return 0;
	}
	vector<int> zeroBits;
	vector<int> oneBits;
	for (int i = 0; i < array.size(); i++) {
		if (fetch(array[i], ind)==1) {
			oneBits.push_back(array[i]);
		} else {
			zeroBits.push_back(array[i]);
		}
	}
	if (zeroBits.size() <= oneBits.size()) {
		int v = findMissing(zeroBits, ind+1);
		return (v<<1) | 0;
	} else {
		int v = findMissing(oneBits, ind+1);
		return (v<<1) | 1;
	}
}


int main() {
	int a[6] = {0, 4, 5, 6, 1, 2};
	vector<int> b;
	for (int i = 0; i < 6; i++) {
		b.push_back(a[i]);
	}
	cout<<findMissing1(6, a)<<endl;
	cout<<findMissing2(b, 0)<<endl;
	return 0;
}