题目
约束
题解
方法一：哈希表
class Solution {
public:
int majorityElement(vector<int>& nums) {
unordered_map<int, int> counts;
int majority = 0, cnt = 0;
for (int num: nums) {
++counts[num];
if (counts[num] > cnt) {
majority = num;
cnt = counts[num];
}
}
return majority;
}
};
方法二：排序
class Solution {
public:
int majorityElement(vector<int>& nums) {
sort(nums.begin(), nums.end());
return nums[nums.size() / 2];
}
};
方法三：随机化
class Solution {
public:
int majorityElement(vector<int>& nums) {
while (true) {
int candidate = nums[rand() % nums.size()];
int count = 0;
for (int num : nums)
if (num == candidate)
++count;
if (count > nums.size() / 2)
return candidate;
}
return -1;
}
};
方法四：分治
class Solution {
int count_in_range(vector<int>& nums, int target, int lo, int hi) {
int count = 0;
for (int i = lo; i <= hi; ++i)
if (nums[i] == target)
++count;
return count;
}
int majority_element_rec(vector<int>& nums, int lo, int hi) {
if (lo == hi)
return nums[lo];
int mid = (lo + hi) / 2;
int left_majority = majority_element_rec(nums, lo, mid);
int right_majority = majority_element_rec(nums, mid + 1, hi);
if (count_in_range(nums, left_majority, lo, hi) > (hi - lo + 1) / 2)
return left_majority;
if (count_in_range(nums, right_majority, lo, hi) > (hi - lo + 1) / 2)
return right_majority;
return -1;
}
public:
int majorityElement(vector<int>& nums) {
return majority_element_rec(nums, 0, nums.size() - 1);
}
};
方法五：Boyer-Moore 投票算法
class Solution {
public:
int majorityElement(vector<int>& nums) {
int candidate = -1;
int count = 0;
for (int num : nums) {
if (num == candidate)
++count;
else if (--count < 0) {
candidate = num;
count = 1;
}
}
return candidate;
}
};