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中等
数组
二分查找

540. 有序数组中的单一元素

English Version

题目描述

给你一个仅由整数组成的有序数组，其中每个元素都会出现两次，唯有一个数只会出现一次。

请你找出并返回只出现一次的那个数。

你设计的解决方案必须满足 O(log n) 时间复杂度和 O(1) 空间复杂度。

 

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

解法

方法一：二分查找

给与的数组是有序的，由此可以使用二分查找，那条件该如何判断呢。

先观察一下线性遍历是如何确定目标的：

for (int i = 0; i < n - 1; i += 2) {
    if (nums[i] != nums[i + 1]) {
        return nums[i];
    }
}
return nums[n - 1];

偶数下标：当 nums[i] != nums[i + 1] && i % 2 == 0 成立，结果便是 nums[i]。 奇数下标：当 nums[i] != nums[i - 1] && i % 2 == 1 成立，结果便是 nums[i - 1]。

于是二分模板就有了：

l = 0
r = n - 1
while l < r
    m = l + (r - l) / 2
    if m % 2 == 0
        if nums[m] == nums[m + 1]
            l = m + 1
        else
            r = m
    else
        if nums[m] == nums[m - 1]
            l = m + 1
        else
            r = m
return nums[l]

Python3

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]

Java

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];
    }
}

C++

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];
    }
};

Go

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]
}

TypeScript

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];
}

Rust

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]
    }
}

C

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];
}