Question

540. Single Element in a Sorted Array

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Description

You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once.

Return _the single element that appears only once_.

Your solution must run in O(log n) time and O(1) space.

 

Example 1:

Input: nums = [1,1,2,3,3,4,4,8,8]
Output: 2

Example 2:

Input: nums = [3,3,7,7,10,11,11]
Output: 10

 

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 105

Solutions

Solution 1

class Solution:
  def singleNonDuplicate(self, nums: List[int]) -> int:
    left, right = 0, len(nums) - 1
    while left < right:
      mid = (left + right) >> 1
      # Equals to: if (mid % 2 == 0 and nums[mid] != nums[mid + 1]) or (mid % 2 == 1 and nums[mid] != nums[mid - 1]):
      if nums[mid] != nums[mid ^ 1]:
        right = mid
      else:
        left = mid + 1
    return nums[left]

class Solution {
  public int singleNonDuplicate(int[] nums) {
    int left = 0, right = nums.length - 1;
    while (left < right) {
      int mid = (left + right) >> 1;
      // if ((mid % 2 == 0 && nums[mid] != nums[mid + 1]) || (mid % 2 == 1 && nums[mid] !=
      // nums[mid - 1])) {
      if (nums[mid] != nums[mid ^ 1]) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return nums[left];
  }
}

class Solution {
public:
  int singleNonDuplicate(vector<int>& nums) {
    int left = 0, right = nums.size() - 1;
    while (left < right) {
      int mid = left + right >> 1;
      if (nums[mid] != nums[mid ^ 1])
        right = mid;
      else
        left = mid + 1;
    }
    return nums[left];
  }
};

func singleNonDuplicate(nums []int) int {
  left, right := 0, len(nums)-1
  for left < right {
    mid := (left + right) >> 1
    if nums[mid] != nums[mid^1] {
      right = mid
    } else {
      left = mid + 1
    }
  }
  return nums[left]
}

function singleNonDuplicate(nums: number[]): number {
  let left = 0,
    right = nums.length - 1;
  while (left < right) {
    const mid = (left + right) >> 1;
    if (nums[mid] != nums[mid ^ 1]) {
      right = mid;
    } else {
      left = mid + 1;
    }
  }
  return nums[left];
}

impl Solution {
  pub fn single_non_duplicate(nums: Vec<i32>) -> i32 {
    let mut l = 0;
    let mut r = nums.len() - 1;
    while l < r {
      let mid = (l + r) >> 1;
      if nums[mid] == nums[mid ^ 1] {
        l = mid + 1;
      } else {
        r = mid;
      }
    }
    nums[l]
  }
}

int singleNonDuplicate(int* nums, int numsSize) {
  int left = 0;
  int right = numsSize - 1;
  while (left < right) {
    int mid = left + (right - left) / 2;
    if (nums[mid] == nums[mid ^ 1]) {
      left = mid + 1;
    } else {
      right = mid;
    }
  }
  return nums[left];
}