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LeetCode 268. Missing Number

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题目

原题在此

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

Example 1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:
Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

Constraints:

  • n == nums.length
  • 1 <= n <= $10^4$
  • 0 <= nums[i] <= n
  • All the numbers of nums are unique.

解析

求和,异或都行.非常简单.

代码

c++/xor

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

java/sum

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public int missingNumber(int[] nums) {
    int len = nums.length;
    int sum = (0+len)*(len+1)/2;
    for(int i=0; i<len; i++)
        sum-=nums[i];
    return sum;
}