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?
- Difficulty: Medium
class Solution {
public:
int missingNumber(vector<int>& nums) {
int n = nums.size();
int expect = (n+1) * n / 2;
for(int i = 0; i < n; i ++)
expect -= nums[i];
return expect;
}
};
本文介绍了一种寻找数组中缺失数值的方法,该方法适用于包含n个不同整数的数组,且这些整数取自0到n范围。通过计算等差数列的和与数组实际元素之差,快速找到缺失的数值。
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