268. Missing Number
Given an array containing n distinct numbers taken from 0, 1, 2, ..., n
, find the one that is missing from the array.
Example 1:
Input: [3,0,1] Output: 2
Example 2:
Input: [9,6,4,2,3,5,7,0,1] Output: 8
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
class Solution { public: int missingNumber(vector<int>& nums) { //bitmap法:数组中所有数其实是0----n序列n个数(缺一个),其中缺的数用n来替代了 int result = nums.size(); //0 1 3 for(int i=0;i<nums.size();i++){ result ^= nums[i]; result ^= i; } return result; } };
//法二求和
class Solution { public: int missingNumber(vector<int>& nums) { //求和公式 s = n*(n+1)/2 int n=nums.size(); //正常序列0--n+1的和 0 1 2 3 int sumComm = (n)*(n+1)/2; //原数组0---n+1缺一个的和 int sum=0; for(int i=0;i<n;i++){ sum += nums[i]; } return sumComm -sum; } };