[leetcode/lintcode 题解] 微软面试题：不重复

给定一个数组 a[]，其中除了2个数，别的数都出现了2次，找到不重复的2个数并返回。

在线评测地址: lintcode领扣

Example 1:

Input: a =[1,2,5,5,6,6]
Output: [1,2]
Explanation:
In addition to the 1 and 2 other numbers appear twice, so return [1, 2]
Example 2:

Input: a=[3,2,7,5,5,7]
Output:[2,3]
Explanation:
In addition to the 2 and 3 other numbers appear twice, so return [2, 3]

【题解】

从头到尾依次异或数组中的每一个数字，得到两个只出现一次的数字的异或结果，在结果数字中找到第一个为1的位的位置，记为第N位，以第N位是不是1为标准把原数组中的数字分成两个子数组，第一个子数组中每个数字的第N位都为1，而第二个子数组的每个数字的第N位都为0。

public class Solution {

    /**
     * @param a: the array
     * @return: the two numbers that are not repeated
     */
    public List<Integer> theTwoNumbers(List<Integer> a) {
        // Write your code here
        int length = a.size();
        List<Integer> res = new ArrayList<Integer>();
        if(length < 2) {
            return res;
        }

        // get num1 ^ num2
        int resultExclusiveOR = 0;
        for(int i = 0; i < length; ++i) {
            resultExclusiveOR ^= a.get(i);
        }

        // get index of the first bit, which is 1 in resultExclusiveOR
        int indexOf1 = FindFirstBitIs1(resultExclusiveOR);

        int num1 = 0, num2 = 0;
        for(int j = 0; j < length; ++j) {
            // divide the numbers in data into two groups,
            // the indexOf1 bit of numbers in the first group is 1,
            // while in the second group is 0
            if(IsBit1(a.get(j), indexOf1)) {
                  num1 ^= a.get(j);
            }
            else {
                  num2 ^= a.get(j);
            }
        }
        res.add(num1);
        res.add(num2);
        return res;
    }

    // Find the index of first bit which is 1 in num (assuming not 0)
    private int FindFirstBitIs1(int num) {
        int indexBit = 0;
        while (((num & 1) == 0) && (indexBit < 32)) {
            num = num >> 1;
            ++ indexBit;
        }
        return indexBit;
    }

    // Is the indexBit bit of num 1?
    private boolean IsBit1(int num, int indexBit) {
        num = num >> indexBit;
        return (num & 1) == 0 ? false : true;
    }
}

更多语言代码参见：leetcode/lintcode题解