给定一个数组,包含从 1 到 N 所有的整数,但其中缺了两个数字。你能在 O(N) 时间内只用 O(1) 的空间找到它们吗?
以任意顺序返回这两个数字均可。
示例 1:
输入: [1]
输出: [2,3]
示例 2:
输入: [2,3]
输出: [1,4]
提示:
nums.length <= 30000
异或运算。与面试题 56 - I. 数组中数字出现的次数 类似。
class Solution:
def missingTwo(self, nums: List[int]) -> List[int]:
res, n = 0, len(nums)
for i in range(n):
res ^= nums[i]
res ^= (i + 1)
res ^= (n + 1)
res ^= (n + 2)
pos = 0
while (res & 1) == 0:
pos += 1
res >>= 1
a = b = 0
for num in nums:
t = num >> pos
if (t & 1) == 0:
a ^= num
else:
b ^= num
for i in range(1, n + 3):
t = i >> pos
if (t & 1) == 0:
a ^= i
else:
b ^= i
return [a, b]
class Solution {
public int[] missingTwo(int[] nums) {
int res = 0, n = nums.length;
for (int i = 0; i < n; ++i) {
res ^= nums[i];
res ^= (i + 1);
}
res ^= (n + 1);
res ^= (n + 2);
int pos = 0;
while ((res & 1) == 0) {
pos += 1;
res >>= 1;
}
int a = 0, b = 0;
for (int num : nums) {
int t = num >> pos;
if ((t & 1) == 0) {
a ^= num;
} else {
b ^= num;
}
}
for (int i = 1; i <= n + 2; ++i) {
int t = i >> pos;
if ((t & 1) == 0) {
a ^= i;
} else {
b ^= i;
}
}
return new int[]{a, b};
}
}