[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.

解題思路：

題意爲找出0~n中缺少的那個數。要求O(n)的時間複雜度和O(1)的空間複雜度。

若採用排序查找的辦法，則需要O(nlogn)的時間。

若採用hash的辦法，則需要O(n)的空間複雜度。

採用位操作的辦法。對數組中數的二進制位的1進行計數。然後對0~n中的二進制位的1進行計數。則兩次計數的差則是missing number貢獻的。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        vector<int> bitCount(sizeof(int) * 8, 0);           //n + 1個數中的bit計數
        vector<int> missingBitCount(sizeof(int) * 8, 0);    //n個數中的bit計數
        
        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;
    }
};