Given an array contains N numbers of 0 .. N, find which number doesn't exist in the array.
Example
Given N = 3 and the array [0, 1, 3], return 2.
Challenge
Do it in-place with O(1) extra memory and O(n) time.
这道题是LeetCode上的原题,请参见我之前的博客Missing Number 丢失的数字。那道题用了两种方法解题,但是LintCode的OJ更加严格,有一个超大的数据集,求和会超过int的范围,所以对于解法一的话需要用long来计算数组之和,其余部分都一样,记得最后把结果转成int即可,参见代码如下:
解法一:
class Solution {
public:
/**
* @param nums: a vector of integers
* @return: an integer
*/
int findMissing(vector
// write your code here
long sum = 0, n = nums.size();
for (auto &a : nums) {
sum += a;
}
return (int)(n * (n + 1) * 0.5 - sum);
}
};
用位操作Bit Manipulation和之前没有区别,参见代码如下:
解法二:
class Solution {
public:
/**
* @param nums: a vector of integers
* @return: an integer
*/
int findMissing(vector
// write your code here
int res = 0;
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size(); ++i) {
res ^= nums[i] ^ (i + 1);
}
return res;
}
};
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