Solutions to high-frequency interview questions of LeetCode in C++17, taking into account both efficiency and comprehensibility.
0
到 nums.size()
和,再减去数组中所有元素即为结果class Solution {
public:
int missingNumber(vector<int>& nums) {
int n = nums.size();
n = n * (n + 1) / 2;
return accumulate(nums.begin(), nums.end(), n, minus<int>{});
}
};
class Solution {
public:
int missingNumber(vector<int>& nums) {
int res = nums.size();
for (int i = 0; i < nums.size(); ++i) {
res += i - nums[i];
}
return res;
}
};