LeetCode Solution-136

136. Single Number

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

思路：
首先将数组排序，然后判断每个数与前后是否都不相等，如果都不相等则返回。注意首尾的值单独拿出来比较。

Solution:

int singleNumber(vector<int>& nums) {
    int n = nums.size();
    if (n == 1) return nums[0];
    sort(nums.begin(), nums.end());
    
    if (nums[0] != nums[1]) return nums[0];
    for (int i = 1; i < n-2; i++) {
        if (nums[i] != nums[i-1] && nums[i] != nums[i+1])
            return nums[i];
    }
    return nums[n-1];
}

性能：
Runtime: 24 ms??Memory Usage: 9.9 MB

LeetCode Solution-136

原文：https://www.cnblogs.com/dysjtu1995/p/12267162.html