[LeetCode] Missing Number

Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

解题思路：

题意为找出0~n中缺少的那个数。要求O(n)的时间复杂度和O(1)的空间复杂度。

若采用排序查找的办法，则需要O(nlogn)的时间。

若采用hash的办法，则需要O(n)的空间复杂度。

采用位操作的办法。对数组中数的二进制位的1进行计数。然后对0~n中的二进制位的1进行计数。则两次计数的差则是missing number贡献的。

class Solution {public:    int missingNumber(vector<int>& nums) {        vector<int> bitCount(sizeof(int) * 8, 0);           //n + 1个数中的bit计数        vector<int> missingBitCount(sizeof(int) * 8, 0);    //n个数中的bit计数                int len = nums.size();        for(int i=0; i<len; i++){            int j = 0;            int num = nums[i];            while(num){                if(num & 1){                    missingBitCount[j]++;                }                j++;                num = num >> 1;            }        }                for(int i = 0; i<=len; i++){            int j = 0;            int num = i;            while(num){                if(num & 1){                    bitCount[j]++;                }                j++;                num = num >> 1;            }        }                int result = 0;        for(int i = sizeof(int) * 8 - 1; i>=0; i--){            result = result << 1;            result += bitCount[i] - missingBitCount[i];        }        return result;    }};