描述：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.
思路1：Moore voting algorithm--每找出两个不同的element，就成对删除即count--，最终剩下的一定就是所求的。时间复杂度：O(n)
class Solution {
public:
int majorityElement(vector<int> &num) {
int elem = ;
int count = ;
for(int i = ; i < num.size(); i++) {
if(count == ) {
elem = num[i];
count = ;
}
else {
if(elem == num[i])
count++;
else
count--;
}
}
return elem;
}
};
思路2：随机挑选一个元素，检查是否是多数元素。时间复杂度：Average：O(n)。期望查找次数 <2
class Solution {
public:
int majorityElement(vector<int> &num) {
int count = ;
for(;;) {
if(num.size() == )
return num[];
else {
int i = rand() % (num.size() - );
for(int j = ; j < num.size(); j++) {
if(num[j] == num[i])
count++;
}
if(count > (num.size() / ))
return num[i];
else {
count = ;
continue;
}
}
}
}
};
附LeetCode建议解决方案：