Given an array of integers, every element appears twice except for one. Find that single one.
Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
题目要求O(n)时间复杂度,O(1)空间复杂度。
思路:利用异或操作。异或的性质1:交换律a ^ b = b ^ a,性质2:0 ^ a = a。于是利用交换律可以将数组假想成相同元素全部相邻,于是将所有元素依次做异或操作,相同元素异或为0,最终剩下的元素就为Single Number。时间复杂度O(n),空间复杂度O(1)
C++:
1 class Solution { 2 public: 3 int singleNumber(vector<int>& nums) { 4 int len = nums.size(), a = 0; 5 for(int i = 0; i < len; i++){ 6 a ^= nums[i]; 7 } 8 return a; 9 } 10 };
c:
1 int singleNumber(int* nums, int numsSize) { 2 int i, a = 0; 3 for(i = 0; i < numsSize; i++){ 4 a ^= nums[i]; 5 } 6 return a; 7 }