var singleNumber = function(nums) {
for(var i = 1,len = nums.length;i<len;i++) {
nums[0] ^= nums[i]
}
return nums[0]
};
獲取數(shù)組中只出現(xiàn)過一次的數(shù)字的算法，用到了XOR，但不太能理解
singleNumber([1,3,1,4,6,4,6,5,3])
[3, 3, 1, 4, 6, 4, 6, 5, 3]
[7, 3, 1, 4, 6, 4, 6, 5, 3]
[1, 3, 1, 4, 6, 4, 6, 5, 3]
[5, 3, 1, 4, 6, 4, 6, 5, 3]
[3, 3, 1, 4, 6, 4, 6, 5, 3]
[6, 3, 1, 4, 6, 4, 6, 5, 3]
[5, 3, 1, 4, 6, 4, 6, 5, 3]
只記得自身的XOR會得到0
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