Question

剑指 Offer II 070. 排序数组中只出现一次的数字

题目描述

给定一个只包含整数的有序数组 nums ，每个元素都会出现两次，唯有一个数只会出现一次，请找出这个唯一的数字。

 

示例 1:

输入: nums = [1,1,2,3,3,4,4,8,8]
输出: 2

示例 2:

输入: nums =  [3,3,7,7,10,11,11]
输出: 10

 

 

提示:

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

 

进阶: 采用的方案可以在 O(log n) 时间复杂度和 O(1) 空间复杂度中运行吗？

 

注意：本题与主站 540 题相同：https://leetcode.cn/problems/single-element-in-a-sorted-array/

解法

方法一

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];
}