Given an integer array of size n,
find all elements that appear more than ⌊ n/3 ⌋
times.
The algorithm should run in linear time and in O(1) space.
若num与n1或n2相同,则将其对应的出现次数加1
否则,若c1或c2为0,则将其置为1,对应的候选众数置为num
否则,将c1与c2分别减1
最后,再统计一次候选众数在数组中出现的次数,若满足要求,则返回之。
class Solution {
public:
vector<int> majorityElement(vector<int>& nums) {
vector<int> res;
int len = nums.size();
if (len ==0){
return res;
}
if (len == 1){
res.push_back(nums[0]);
return res;
}
int m1=nums[0], cnt1 = 1;
int m2=0, cnt2 = 0;
for (int i = 1; i < len; i++){
if (nums[i] == m1){//顺序不能乱
cnt1++;
}
else if (nums[i] == m2){
cnt2++;
}
else if (cnt1 == 0){
m1 = nums[i];
cnt1 = 1;
}
else if (cnt2 == 0){
m2 = nums[i];
cnt2 = 1;
}
else{
cnt1--;
cnt2--;
}
}
cnt1 = 0, cnt2 = 0;
for (int i = 0; i < len; i++){
if (nums[i] == m1){
cnt1++;
}
if (nums[i] == m2){
cnt2++;
}
}
if (cnt1>len / 3){
res.push_back(m1);
}
if (cnt2 > len / 3){
if (m2!=m1)
res.push_back(m2);
}
return res;
}
};