代码拉取完成，页面将自动刷新
同步操作将从 doocs/leetcode 强制同步，此操作会覆盖自 Fork 仓库以来所做的任何修改，且无法恢复！！！
确定后同步将在后台操作，完成时将刷新页面，请耐心等待。
You have a set of integers s, which originally contains all the numbers from 1 to n. Unfortunately, due to some error, one of the numbers in s got duplicated to another number in the set, which results in repetition of one number and loss of another number.
You are given an integer array nums representing the data status of this set after the error.
Find the number that occurs twice and the number that is missing and return them in the form of an array.
Example 1:
Input: nums = [1,2,2,4] Output: [2,3]
Example 2:
Input: nums = [1,1] Output: [1,2]
Constraints:
2 <= nums.length <= 1041 <= nums[i] <= 104class Solution:
def findErrorNums(self, nums: List[int]) -> List[int]:
res = 0
for num in nums:
res ^= num
for i in range(1, len(nums) + 1):
res ^= i
pos = 0
while (res & 1) == 0:
res >>= 1
pos += 1
a = b = 0
for num in nums:
if ((num >> pos) & 1) == 0:
a ^= num
else:
b ^= num
for i in range(1, len(nums) + 1):
if ((i >> pos) & 1) == 0:
a ^= i
else:
b ^= i
for num in nums:
if num == a:
return [a, b]
return [b, a]
class Solution {
public int[] findErrorNums(int[] nums) {
int res = 0;
for (int num : nums) {
res ^= num;
}
for (int i = 1, n = nums.length; i < n + 1; ++i) {
res ^= i;
}
int pos = 0;
while ((res & 1) == 0) {
res >>= 1;
++pos;
}
int a = 0, b = 0;
for (int num : nums) {
if (((num >> pos) & 1) == 0) {
a ^= num;
} else {
b ^= num;
}
}
for (int i = 1, n = nums.length; i < n + 1; ++i) {
if (((i >> pos) & 1) == 0) {
a ^= i;
} else {
b ^= i;
}
}
for (int num : nums) {
if (num == a) {
return new int[]{a, b};
}
}
return new int[]{b, a};
}
}
此处可能存在不合适展示的内容，页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容，可点击提交进行申诉，我们将尽快为您处理。
马建仓 AI 助手
