给定一个只包含整数的有序数组,每个元素都会出现两次,唯有一个数只会出现一次,找出这个数。
示例 1:
输入: [1,1,2,3,3,4,4,8,8]
输出: 2
示例 2:
输入: [3,3,7,7,10,11,11]
输出: 10
注意: 您的方案应该在 O(log n)时间复杂度和 O(1)空间复杂度中运行。
详见:https://leetcode.com/problems/single-element-in-a-sorted-array/description/
C++:
方法一:
class Solution {
public:
int singleNonDuplicate(vector<int>& nums) {
int res=0;
for(int num:nums)
{
res^=num;
}
return res;
}
};
方法二:
class Solution {
public:
int singleNonDuplicate(vector<int>& nums)
{
int left = 0, right = nums.size() - 1, n = nums.size();
while (left < right)
{
int mid = left + (right - left) / 2;
if (nums[mid] == nums[mid + 1])
{
if ((n - 1 - mid) % 2 == 1)
{
right = mid;
}
else
{
left = mid + 1;
}
}
else
{
if (mid == 0 || nums[mid] != nums[mid - 1])
{
return nums[mid];
}
if ((n - 1 - mid) % 2 == 0)
{
right = mid;
}
else
{
left = mid + 1;
}
}
}
return nums[left];
}
};
参考:http://www.cnblogs.com/grandyang/p/7679036.html