136. Single Number

最新推荐文章于 2025-08-23 17:53:17 发布

weixin_30639719

最新推荐文章于 2025-08-23 17:53:17 发布

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文章标签： runtime python java

原文链接：http://www.cnblogs.com/ruruozhenhao/p/9960655.html

本文介绍了一种在整数数组中找出唯一出现一次元素的算法，该算法具备线性时间复杂度，并提供了C++、Java和Python三种实现方式。文章强调了算法的高效性和内存使用优化，通过示例展示了算法的具体应用。

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Given a non-empty array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

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

Example 2:

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

Approach #1: C++.

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        unordered_map<int, int> mp;
        for (int i = 0; i < nums.size(); ++i) 
            mp[nums[i]]++;
        for (int i = 0; i < nums.size(); ++i)
            if (mp[nums[i]] == 1) return nums[i];
    }
};

Approach #2: Java.

class Solution {
    public int singleNumber(int[] nums) {
        int result = 0;
        for (int i : nums)
            result ^= i;
        return result;
    }
}

Note: ^ return 0 if two elements is same.

Approach #3: Python.

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        dic = {}
        for num in nums:
            dic[num] = dic.get(num, 0) + 1
        for key, val in dic.items():
            if val == 1:
                return key