【第三轮】7.24-7.31 CareerCup 5.7

5.7 An array A contains all the integers from 0 through 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 thejth bit ofAfi]," which takes constant time. Write code to find the missing integer. Can you do it in 0(n) time?


回复解法可以按照以下格式来

【解题思路】
【时间复杂度】
【空间复杂度】
【gist link】
---------------OPTional，如果觉得test case比较好，欢迎写出来分享----------------------
【test case】

Notice：
1、记得在程序注释中表明自己算法的时间、空间复杂度
2、代码难懂之处加注释
3、每道题目有对应的帖子，除了贴解法，欢迎讨论，集思广益
4、任何未尽之处，欢迎回报名帖提问，我会进一步作出修改

【解题思路】这道题不错，想了一阵才有思路
In position k, from n % 2^(k+1) to calculate how much zeros suppose to be more than ones if the actual difference is larger than our result, that means we lost a 1, then we add a 1 to result in position k. Otherwise we lost a zero, we just do nothing.
【时间复杂度】O(N)
【空间复杂度】O(1)
【gist link】https://gist.github.com/Noahsark/d3ca3235e815389211b6

【解题思路】
refer to the solution in the book

【时间复杂度】
O(n)

【空间复杂度】
O(logn)

【gist link】
https://gist.github.com/happyWinner/af5502e05d0e174639bf

【解题思路】
Starting from the least significant bit, check bits for all numbers in the array:
1. Decide whether the missing number has 0 or 1 in this bit position (by checking number of 0s and 1s in all array integers)

2. Recursively call this function with all the numbers has same bits in that position as the missing number, until all the bits(32 or 64 depends on length of Integer)

3. Reforming the missing numbers by left shifting the result of sub routine and appending 0 or 1 at then end.

【时间复杂度】
O(N)

【空间复杂度】
O(N)

【gist link】
https://gist.github.com/chrislukkk/3e676ac9bfbc8a9ddb43

【解题思路】
1. search from LSB, count the number of 1's and 0's, recursively search for the missing number and appendix 0 or 1 after the previous result, depending on relationship between c0 and c1.
2. same idea, using iteration instead
【时间复杂度】
O(n)
【空间复杂度】
O(N)
【gist link】
https://gist.github.com/dd204c15820dbc436bc0

【解题思路】
1. search from LSB, count the number of 1's and 0's, recursively search for the missing number and appendix 0 or 1 after the previous result, depending on relationship between c0 and c1.
2. same idea, using iteration instead
【时间复杂度】
O(n)
【空间复杂度】
O(N)
【gist link】
https://gist.github.com/dd204c15820dbc436bc0

【解题思路】
1. for digit i: count 0s and 1s at that digit, if #0>#1, missing a 1,  delete all numbers with 0 at that digits and do this again for other digits
2. recursive solution
【时间复杂度】
O(N)
【空间复杂度】
O(N)
【gist link】 https://gist.github.com/jyhjuzi/a97d5c52083c9438d147

【解题思路】
参考书上

【时间复杂度】
O(N)

【空间复杂度】
O(N)

【gist link】
https://gist.github.com/UncleGarden/88967999c234cd171775