169. Majority Element--寻找数组中出现次数超过一半的数据，229. Majority Element II--注意最后的检测

第一题、169. Majority Element-

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

方法一、由于出现一半的数字，则即寻找第middle大的数字，可以基于快速排序的partition的思想，然后找出第Middle大的数字，代码此处省略；

方法二、hash的方法；

方法三、基于出现次数超过一半的数据的特征，代码如下：

int majorityElement(vector<int>& nums) {
if(nums.size()<=0)
return -1;

int count = 1;
int result = nums[0];
for(int i = 1; i < nums.size(); i++){
if(count == 0){
result = nums[i];
count = 1;
}
else if(nums[i] == result){
count++;
}
else{
count--;
}

}

return result;
}

第二题、229. Majority Element II

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.

/*求众数的问题
观察可知，数组中至多可能会有2个出现次数超过 ⌊ n/3 ⌋ 的众数;
记变量num1, num2为候选众数； cnum1, cnum2为它们对应的出现次数
遍历数组，记当前数字为num
若num与num1或num2相同，则将其对应的出现次数加1
否则，若cnum1或cnum2为0，则将其置为1，对应的候选众数置为num
否则，将cnum1与cnum2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。*/
vector<int> majorityElement(vector<int>& nums) {
int len = nums.size();
if(len <= 0){
return (vector<int> ());
}

int num1 = 0,num2 = 1;
int cnum1 = 0,cnum2 = 0;
for(int i = 0; i < len; i++){
if(nums[i] == num1){
cnum1++;
}
else if(nums[i] == num2){
cnum2++;
}
else if(cnum1 == 0){
num1 = nums[i];
cnum1 = 1;
}
else if(cnum2 == 0){
num2 = nums[i];
cnum2 = 1;
}
else{
cnum1--;
cnum2--;
}
}

cnum1 = 0;
cnum2 = 0;
for(int i = 0; i < len; i++){
if(nums[i] == num1){
cnum1++;
}
else if(nums[i] == num2){
cnum2++;
}
}

vector<int> result;
if(cnum1 > len / 3){
result.push_back(num1);
}

if(cnum2 > len / 3){
result.push_back(num2);
}

return result;
}