leetcode -----Majority Elemen II
2016-07-07 16:34
204 查看
思路: 摩尔投票法 (一个数组中大于1/3 次数的最多有2个!!!!)
遍历两遍数组,第一遍找到两个候选的众数,第二遍计算这两个数的次数,最后比较时候大于n/3。
vector<int> majorityElement(vector<int>& nums)
{
vector<int> res;
int m=0,n=0,cm=0,cn=0;
for(int i=0;i<nums.size();i++)
{
if(m==nums[i])
cm++;
else if(n==nums[i])
cn++;
else if(cm==0)
{
m=nums[i];
cm=1;
}
else if(cn==0)
{
n=nums[i];
cn=1;
}
else
{
cn--;
cm--;
}
}
cm=0,cn=0;
for(int i=0;i<nums.size();i++)
{
if(nums[i]==m) cm++;
else if(nums[i]==n) cn++;
}
if(cm>nums.size()/3) res.push_back(m);
if(cn>nums.size()/3) res.push_back(n);
return res;
}
遍历两遍数组,第一遍找到两个候选的众数,第二遍计算这两个数的次数,最后比较时候大于n/3。
vector<int> majorityElement(vector<int>& nums)
{
vector<int> res;
int m=0,n=0,cm=0,cn=0;
for(int i=0;i<nums.size();i++)
{
if(m==nums[i])
cm++;
else if(n==nums[i])
cn++;
else if(cm==0)
{
m=nums[i];
cm=1;
}
else if(cn==0)
{
n=nums[i];
cn=1;
}
else
{
cn--;
cm--;
}
}
cm=0,cn=0;
for(int i=0;i<nums.size();i++)
{
if(nums[i]==m) cm++;
else if(nums[i]==n) cn++;
}
if(cm>nums.size()/3) res.push_back(m);
if(cn>nums.size()/3) res.push_back(n);
return res;
}
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