LintCode 1112: Set Mismatch (XOR经典题)

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本文介绍了一种高效算法，用于在包含从1到n的整数集合中，找到重复出现的数字及缺失的数字。通过排序和XOR运算，算法能够准确地定位错误的数字并找出缺失的部分。

Set Mismatch
The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.

Given an array nums representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.

Example
Input: nums = [1,2,2,4]
Output: [2,3]
Notice
1.The given array size will in the range [2, 10000].
2.The given array’s numbers won’t have any order.

我的思路是：
先排序，然后找出nums[i-1]=nums[i]，则num[i]就是nums[]里面出错的那个，设其为candidate_err。
那怎么找到[1…n]里面出错的那个呢？假设其为candidiate_orig。
用xor就可以了。
假设nums = [3,2,3,4,6,5]，排序后为[2,3,3,4,5,6]。
可得candidate_err = 3。
然后XOR所有的nums，再XOR所有的1…n，再XOR candidate_err。最后结果就是candidate_orig = 1。
即 2 ^ 3 ^ 3 ^ 4 ^ 5 ^ 6 ^ 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 3 = 1.
最后还异或3是为了确保3有偶数个，这样3的异或结果就为0了。

代码如下：

class Solution {
public:
    /**
     * @param nums: an array
     * @return: the number occurs twice and the number that is missing
     */
    vector<int> findErrorNums(vector<int> &nums) {
        int len = nums.size();
        if (len == 0) return vector<int>(2, 0);
        
        sort(nums.begin(), nums.end());
        int candidate_orig = 0, candidate_err = 0;
        for (int i = 1; i <= len; ++i) {
            if (nums[i - 1] == nums[i]) {
                candidate_err = nums[i];
                break;
            }
        }
        int xor_result = 0;
        for (int i = 1; i <= len; ++i) {
            xor_result ^= i;
            xor_result ^= nums[i - 1];
        }
        xor_result ^= candidate_err;
        
        candidate_orig = xor_result;
        
        return vector<int>{candidate_err, candidate_orig};
    }
};