有序数组中的单一元素

540. 有序数组中的单一元素

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

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class Solution { public: int singleNonDuplicate(vector<int>& nums) { } };

cpp 解法, 执行用时: 0 ms, 内存消耗: 24.4 MB, 提交时间: 2024-11-10 00:28:13

class Solution {
public:
    int singleNonDuplicate2(vector<int>& nums) {
        int low = 0, high = nums.size() - 1;
        while (low < high) {
            int mid = (high - low) / 2 + low;
            if (nums[mid] == nums[mid ^ 1]) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return nums[low];
    }

    int singleNonDuplicate(vector<int>& nums) {
        int low = 0, high = nums.size() - 1;
        while (low < high) {
            int mid = (high - low) / 2 + low;
            mid -= mid & 1;
            if (nums[mid] == nums[mid + 1]) {
                low = mid + 2;
            } else {
                high = mid;
            }
        }
        return nums[low];
    }
};

golang 解法, 执行用时: 0 ms, 内存消耗: 9.2 MB, 提交时间: 2024-11-10 00:27:37

func singleNonDuplicate2(nums []int) int {
    low, high := 0, len(nums)-1
    for low < high {
        mid := low + (high-low)/2
        mid -= mid & 1
        if nums[mid] == nums[mid+1] {
            low = mid + 2
        } else {
            high = mid
        }
    }
    return nums[low]
}

func singleNonDuplicate(nums []int) int {
    low, high := 0, len(nums)-1
    for low < high {
        mid := low + (high-low)/2
        if nums[mid] == nums[mid^1] {
            low = mid + 1
        } else {
            high = mid
        }
    }
    return nums[low]
}

java 解法, 执行用时: 0 ms, 内存消耗: 48.9 MB, 提交时间: 2024-11-10 00:27:09

class Solution {
    public int singleNonDuplicate(int[] nums) {
        int low = 0, high = nums.length - 1;
        while (low < high) {
            int mid = (high - low) / 2 + low;
            if (nums[mid] == nums[mid ^ 1]) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return nums[low];
    }

    public int singleNonDuplicate2(int[] nums) {
        int low = 0, high = nums.length - 1;
        while (low < high) {
            int mid = (high - low) / 2 + low;
            mid -= mid & 1;
            if (nums[mid] == nums[mid + 1]) {
                low = mid + 2;
            } else {
                high = mid;
            }
        }
        return nums[low];
    }
}

python3 解法, 执行用时: 44 ms, 内存消耗: 21.2 MB, 提交时间: 2022-11-21 11:25:44

class Solution:
    def singleNonDuplicate(self, nums: List[int]) -> int:
        low, high = 0, len(nums) - 1
        while low < high:
            mid = (low + high) // 2
            mid -= mid & 1
            if nums[mid] == nums[mid + 1]:
                low = mid + 2
            else:
                high = mid
        return nums[low]

python3 解法, 执行用时: 36 ms, 内存消耗: 21.2 MB, 提交时间: 2022-11-21 11:24:12

class Solution:
    def singleNonDuplicate(self, nums: List[int]) -> int:
        low, high = 0, len(nums) - 1
        while low < high:
            mid = (low + high) // 2
            if nums[mid] == nums[mid ^ 1]: # mid为偶数，则+1，反之-1
                low = mid + 1
            else:
                high = mid
        return nums[low]

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