[LeetCode] 169. Majority Element 多数元素

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.

Example 1:

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

Example 2:

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

给一个大小为n的数组，找出它的多数元素。数组非空，多数元素存在。多数元素：出现次数超过n/2的元素(超过数组长度一半)。

解法1: 用两个循环跟踪统计最多次数的元素。 T：O(n*n)  S: O(1)

解法2: BST,  T：O(nlogn)  S: O(n)

解法3：Boyer-Moore多数投票算法 Boyer–Moore majority vote algorithm，T：O(n)  S: O(1)

解法4: HashMap， T：O(n)  S: O(n)

Java:

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

public class Solution {     public int majorityElement(int[] num) {           int major=num[0], count = 1;         for(int i=1; i<num.length;i++){             if(count==0){                 count++;                 major=num[i];             }else if(major==num[i]){                 count++;             }else count--;                       }         return major;     } }

Java：

?

1 2 3 4 5 6 7 8 9 10 11

public class Solution {     public int majorityElement(int[] nums) {         int res = 0, cnt = 0;         for (int num : nums) {             if (cnt == 0) {res = num; ++cnt;}             else if (num == res) ++cnt;             else --cnt;         }         return res;     } }

Python: T: O(n), S: O(1)

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14

class Solution:     def majorityElement(self, nums):         idx, cnt = 0, 1                   for i in xrange(1, len(nums)):             if nums[idx] == nums[i]:                 cnt += 1             else:                 cnt -= 1                 if cnt == 0:                     idx = i                     cnt = 1                   return nums[idx]

C++:

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

class Solution { public:     int majorityElement(vector<int>& nums) {         int ans = nums[0], cnt = 1;         for (const auto& i : nums) {             if (i == ans) {                 ++cnt;             } else {                 --cnt;                 if (cnt == 0) {                     ans = i;                     cnt = 1;                 }             }         }         return ans;     } };

 

类似题目：

[LeetCode] 229. Majority Element II  多数元素 II

 

All LeetCode Questions List 题目汇总