Missing Number
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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?
解法一:
最简单的就是开个n+1的bool数组,然后遍历一下输入数组,存在则置为true。然后再查找false所对应的值,但这样的话虽然线性时间完成,但需要额外开一个大小为n的数组,虽然能AC,但不符合规则。
解法二:
我们知道前0到n的和为n(n+1)/2,因此我们可以将数组中的整数相加得到k,然后答案就是n(n+1)/2
class Solution {public: int missingNumber(vector<int>& nums) { int res=0; for(int i=0;i<nums.size();++i)res+=nums[i]; return (nums.size()*(nums.size()+1))/2-res; }};
0 0
- Missing number
- Missing number
- Missing number
- Missing number
- Missing Number
- Missing Number
- Missing Number
- Missing Number.
- Missing Number
- Missing Number
- Missing Number
- Missing Number
- Missing Number
- Missing Number
- Missing Number
- Missing Number
- Missing Number
- Missing Number
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