You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:

Input: nums = [2,2,1]
Output: 1

Example 2:

Input: nums = [4,1,2,1,2]
Output: 4

Example 3:

Input: nums = [1]
Output: 1

Constraints:

• 1 <= nums.length <= 3 * 104
• -3 * 104 <= nums[i] <= 3 * 104
• Each element in the array appears twice except for one element which appears only once.

Solution: XOR

single_number ^ a ^ b ^ c ^ … ^ a ^ b ^ c … = single_number

Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
  int singleNumber(vector<int>& nums) {
    int ans = 0;
    for (int x : nums)
      ans ^= x;
    return ans;
  }
};

Related Problems

• 花花酱 LeetCode 137. Single Number II

The post 花花酱 LeetCode 136. Single Number appeared first on Huahua's Tech Road.

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Example 1:

Input: [1,3,4,2,2]
Output: 2

Example 2:

Input: [3,1,3,4,2]
Output: 3

Note:

1. You must not modify the array (assume the array is read only).
2. You must use only constant, O(1) extra space.
3. Your runtime complexity should be less than O(n²).
4. There is only one duplicate number in the array, but it could be repeated more than once.

Solution1: Binary Search

Time complexity: O(nlogn)

Space complexity: O(1)

Find the smallest m such that len(nums <= m) > m, which means m is the duplicate number.

In the sorted form [1, 2, …, m1, m2, m + 1, …, n]

There are m+1 numbers <= m

// Author: Huahua
// Running time: 8 ms
class Solution {
public:
  int findDuplicate(vector<int>& nums) {
    int l = 1;
    int r = nums.size();
    while (l < r) {
      int m = (r - l) / 2 + l;
      int count = 0; // len(nums <= m)
      for (int num : nums)
        if (num <= m) ++count;
      if (count <= m)
        l = m + 1;
      else
        r = m;
    }
    return l;
  }
};

Solution: Linked list cycle

Convert the problem to find the entry point of the cycle in a linked list.

Take the number in the array as the index of next node.

[1,3,4,2,2]

0->1->3->2->4->2 cycle: 2->4->2

[3,1,3,4,2]

0->3->4->2->3->4->2 cycle 3->4->2->3

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  int findDuplicate(vector<int>& nums) {
    int slow = 0;
    int fast = 0;
    while (true) {
      slow = nums[slow];
      fast = nums[nums[fast]];
      if (slow == fast) break;
    }
    fast = 0;
    while (fast != slow) {
      slow = nums[slow];
      fast = nums[fast];
    }
    return slow;
  }
};

The post 花花酱 LeetCode 287. Find the Duplicate Number appeared first on Huahua's Tech Road.