136. Single Number

Given an array of integers, every element appearstwice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

使用map：

public  int singleNumber(int[] nums) {
		int i = 0;
		int j = nums.length - 1;
		Map<Integer, Integer> map = new HashMap<>();
		while (i <= j) {
			if (map.containsKey(nums[i]))
				map.put(nums[i], map.get(nums[i])+1);
			else
				map.put(nums[i], 1);
			i++;
			
		}
		i = 0;
		while (i <= j) {
			if (map.get(nums[i]) == 1)
				break;
			i++;
		}
		return nums[i];
	}
 

如果a、b兩個值不相同，則異或結果為1。如果a、b兩個值相同，異或結果為0。

0與任何數做異或都是 那個數，相同的數字做異或等於0；

    1. a ^ b = b ^ a

    2. a ^ b ^ c = a ^ （b ^ c） = （a ^ b） ^ c;

public  int singleNumber(int[] nums) {

		int result = nums[0];

		for (int i = 1; i < nums.length; i++) {
			result = result ^ nums[i];
		}
		
		return result;
	}