[LeetCode] 136. Single Number

There are two ways for the problem:
1. Tradition: Setup a map or set.
2. Bit operation: x ^ x = 0; so if the number keep XOR, the same ones will cancel each other, the one that left would be the single number.

http://www.programcreek.com/2012/12/leetcode-solution-of-single-number-in-java/

这道题需要注意异或的几个知识点:

• a^0 = a
• a^a = 0
• a^b = b^a
• a^b^c = a^(b^c)
• 得到一个数二进制最右位为1，其余为0的模：a &(~a+1)   (~a + 1) = -a  =>这道题用不上，但这个知识点很有用。

 

public class Solution {
    /**
      *@param A : an integer array
      *return : a integer 
      */
    public int singleNumber(int[] A) {
        // Write your code here
        Map<Integer, Integer> base = new HashMap<Integer, Integer>();
        for(int a : A){
            if(base.containsKey(a)){
                base.put(a, 1);
            } else {
                base.put(a, 0);
            }
        }
        
        int result = 0;
        for (Map.Entry<Integer, Integer> entry : base.entrySet()){
            if (entry.getValue() == 0){
                result = entry.getKey();
                break;
            }
        }
        return result;
    }
}

　　

class Solution {
    public int singleNumber(int[] nums) {
        if (nums == null || nums.length == 0) {
            return Integer.MIN_VALUE;
        }
        int eor = 0;
        for (int i = 0; i < nums.length; i++) {
            eor ^= nums[i];
        }
        return eor;
    }
}