Leetcode 268. Missing Number

268. Missing Number

• Total Accepted: 62408
• Total Submissions: 150160
• Difficulty: Medium

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?

 

 

思路：先排序再查找。

 

代码：

方法一：a^b^b=a以及异或操作的交换律。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int i,Xor=0;
        for(i=0;i<nums.size();i++){
            Xor=Xor^i^nums[i];
        }
        return Xor^i;
    }
};

 

方法二：对[0,n]求和后减去sum(nums[0],...,nums[n-1])。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int i,sum=nums.size();
        for(i=0;i<nums.size();i++){
            sum+=i-nums[i];
        }
        return sum;
    }
};

 

方法三：二分查找。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int mid,left=0,right=nums.size()-1;
        while(left<=right){
            mid=left+(right-left)/2;
            if(nums[mid]>mid){
                right=mid-1;
            }else{
                left=mid+1;
            }
        }
        return left;
    }
};

 

方法四：

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int i=0,res=1;
        sort(nums.begin(),nums.end());
        while(i<nums.size()&&(res&1)^(nums[i]&1)){
            res=nums[i++];
        }
        return i<nums.size()?nums[i]-1:i;
    }
};