linear-list/array/majority-element

Majority Element

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.

分析

这题最简单的解法，先把数组排序，O(nlogn)，然后从头到尾扫描一遍，找出最长的连续子串。

由于超过⌊ n/2 ⌋次，可以对上面的方法改进一下，排序后，不需要扫描，直接返回中间那个元素，nums[n/2]，肯定就是答案。

上述两个方法都是 O(nlogn)的，本题有一个线性解法。由于超过⌊ n/2 ⌋，可以用相抵消的思想，凡是和最多元素不相等的，就抵消，最后剩下的一定就是最多的那个元素。

代码

解法 1 排序

# Majority Element
# Time Complexity: O(nlogn), Space Complexity: O(1)
class Solution:
    def majorityElement(self, nums):
        nums.sort()
        return nums[len(nums)//2]

// Majority Element
// Time Complexity: O(nlogn), Space Complexity: O(1)
public class Solution {
    public int majorityElement(int[] nums) {
        Arrays.sort(nums);
        return nums[nums.length/2];
    }
}

// Majority Element
// Time Complexity: O(nlogn), Space Complexity: O(1)
class Solution {
public:
    int majorityElement(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        return nums[nums.size()/2];
    }
};

解法 2 Boyer-Moore Voting Algorithm

# Majority Element
# Time Complexity: O(nlogn), Space Complexity: O(1)
class Solution:
    def majorityElement(self, nums):
        candidate = 0
        count = 0

        for num in nums:
            if candidate == num:
                count += 1
            elif count == 0:
                candidate = num
                count = 1
            else:
                count -= 1

        return candidate

// Majority Element
// Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
    public int majorityElement(int[] nums) {
        int count = 0;
        int candidate = 0;

        for (int num : nums) {
            if (candidate == num) {
                ++count;
            } else if (count == 0) {
                candidate = num;
                count = 1;
            } else {
                --count;
            }
        }
        return candidate;
    }
}

// Majority Element
// Time Complexity: O(nlogn), Space Complexity: O(1)
class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int count = 0;
        int candidate = 0;

        for (int num : nums) {
            if (candidate == num) {
                ++count;
            } else if (count == 0) {
                candidate = num;
                count = 1;
            } else {
                --count;
            }
        }
        return candidate;
    }
};

相关题目

• Majority Element II