C++实现LeetCode(136.单独的数字)

[LeetCode] 136.Single Number 单独的数字

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

这道题给了我们一个非空的整数数组,说是除了一个数字之外所有的数字都正好出现了两次,让我们找出这个只出现一次的数字。题目中让我们在线性的时间复杂度内求解,那么一个非常直接的思路就是使用 HashSet,利用其常数级的查找速度。遍历数组中的每个数字,若当前数字已经在 HashSet 中了,则将 HashSet 中的该数字移除,否则就加入 HashSet。这相当于两两抵消了,最终凡事出现两次的数字都被移除了 HashSet,唯一剩下的那个就是单独数字了,参见代码如下:

C++ 解法一:

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        unordered_set<int> st;
        for (int num : nums) {
            if (st.count(num)) st.erase(num);
            else st.insert(num);
        }
        return *st.begin();
    }
};

Java 解法一:

class Solution {
    public int singleNumber(int[] nums) {
        Set<Integer> st = new HashSet<>();
        for (int num : nums) {
            if (!st.add(num)) st.remove(num);
        }
        return st.iterator().next();
    }
}

题目中让我们不使用额外空间来做,本来是一道非常简单的题,但是由于加上了时间复杂度必须是 O(n),并且空间复杂度为 O(1),使得不能用排序方法,也不能使用 HashSet 数据结构。那么只能另辟蹊径,需要用位操作 Bit Operation 来解此题,这个解法如果让我想,肯定想不出来,因为谁会想到用逻辑异或来解题呢。逻辑异或的真值表为:

 异或运算的真值表如下:

A B
F F F
F T T
T F T
T T F

由于数字在计算机是以二进制存储的,每位上都是0或1,如果我们把两个相同的数字异或,0与0 '异或' 是0,1与1 '异或' 也是0,那么我们会得到0。根据这个特点,我们把数组中所有的数字都 '异或' 起来,则每对相同的数字都会得0,然后最后剩下来的数字就是那个只有1次的数字。这个方法确实很赞,但是感觉一般人不会往 '异或' 上想,绝对是为CS专业的同学设计的好题呀,赞一个~~ 

C++ 解法二:

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int res = 0;
        for (auto num : nums) res ^= num;
        return res;
    }
};

Java 解法二:

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

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