leetcode268：Missing Number

描写叙述

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

方法一：时间复杂度O(n),空间复杂度O(n/32)

传统方法， 同一时候使用位运算压缩空间（int型32位标记32个数字）

//36ms
class Solution {
public:
int missingNumber(vector<int>& nums) {
int n = nums.size() + 1;
int c = (n >> 5) + 1;
int *p = new int[c];
memset(p, 0, c*sizeof(int));
for (int x : nums) p[x >> 5] |= (1 << x % 32);
for (int i = 0; i<n; i++) if(((p[i >> 5] >> (i % 32)) & 1) == 0) return i;
}
};

方法二：

位运算,将n+1个数字拓展到2^(lg(n)+1)个数。所有异或就可以得到缺失的数字。

时间复杂度O(2^(lg(n)+1))，空间复杂度O(1)

//36ms
class Solution {
public:
int missingNumber(vector<int>& nums) {
int n = nums.size() + 1;
int ans = 0;
int _pow = (int)pow(2, 1 + (int)(log(n) / log(2)));
for (int x : nums) ans ^= x;
for (int i = n; i < _pow; i++) ans ^= i;
return ans;
}
};