题目:
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?
解析:
1 class Solution { 2 public: 3 int missingNumber(vector<int>& nums) { 4 sort(nums.begin(),nums.end()); 5 for(int i = 0; i <= nums.size(); i++) 6 if(nums[i] != i) 7 return i; 8 } 9 };
高效的算法应该是使用位运算,我没明白,也不想深究:
class Solution { public: int missingNumber(vector<int>& nums) { int result = 0; for (int i = 0; i < nums.size(); i++) result ^= nums[i]^(i+1); return result; } };