LeetCode---(136) Single Number
2015-05-18 15:07
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Given an array of integers, every element appears twice except
for one. Find that single one.
常规的做法,就是把数组排序,然后进行比较
class Solution {
public:
int singleNumber(vector<int>& nums) {
int N=nums.size();
if(N==1)
return nums[0];
sort(nums.begin(),nums.end());
for(int i=0;i<N;i++)
{
if(i%2==0)
{
if(nums[i]!=nums[i+1])
{
return nums[i];
break;
}
}
}
}
};
也可以巧用异或运算符,异或,不仅能处理两次的情况,只要出现偶数次,都可以清零。
class Solution {
public:
int singleNumber(vector<int>& nums) {
int N=nums.size();
int res=0;
for(int i=0;i<N;i++)
{
res^=nums[i];
}
return res;
}
};
for one. Find that single one.
常规的做法,就是把数组排序,然后进行比较
class Solution {
public:
int singleNumber(vector<int>& nums) {
int N=nums.size();
if(N==1)
return nums[0];
sort(nums.begin(),nums.end());
for(int i=0;i<N;i++)
{
if(i%2==0)
{
if(nums[i]!=nums[i+1])
{
return nums[i];
break;
}
}
}
}
};
也可以巧用异或运算符,异或,不仅能处理两次的情况,只要出现偶数次,都可以清零。
class Solution {
public:
int singleNumber(vector<int>& nums) {
int N=nums.size();
int res=0;
for(int i=0;i<N;i++)
{
res^=nums[i];
}
return res;
}
};
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