[LeetCode] 169. Majority Element

[frdmu]

Given an array nums of size n, return the majority element.

The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.

 

Example 1:

Input: nums = [3,2,3]
Output: 3

Example 2:

Input: nums = [2,2,1,1,1,2,2]
Output: 2

Constraints:

• n == nums.length
• 1 <= n <= 5 * 104
• -231 <= nums[i] <= 231 - 1
   

Follow-up: Could you solve the problem in linear time and in O(1) space?

 
 
 
求众数，如果使空间复杂度是O(1)，用摩尔计数法：
代码如下：

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int candidate = nums[0], cnt = 0, n = nums.size();
        
        for (auto e: nums) {
            if (candidate == e) {
                cnt++;
            } else {
                if (cnt == 0) {
                    candidate = e;
                    cnt++;
                } else {
                    cnt--;
                }
            }
        }
        cnt = 0;
        for (auto e: nums) {
            if (candidate == e)
                cnt++;        
        }
        
        if (cnt > n / 2)
            return candidate;
        return -1;
    }
};

类似题目：
229. Majority Element II