Majority Number

最后更新于:2022-04-02 01:09:56

# Majority Number ### Source - leetcode: [Majority Element | LeetCode OJ](https://leetcode.com/problems/majority-element/) - lintcode: [(46) Majority Number](http://www.lintcode.com/en/problem/majority-number/) ~~~ Given an array of integers, the majority number is the number that occurs more than half of the size of the array. Find it. Example Given [1, 1, 1, 1, 2, 2, 2], return 1 Challenge O(n) time and O(1) extra space ~~~ ### 题解 找出现次数超过一半的数,使用哈希表统计不同数字出现的次数,超过二分之一即返回当前数字。这种方法非常简单且容易实现,但会占据过多空间,注意到题中明确表明要找的数会超过二分之一,这里的隐含条件不是那么容易应用。既然某个数超过二分之一,那么用这个数和其他数进行 PK,不同的计数器都减一**(核心在于两两抵消)**,相同的则加1,最后返回计数器大于0的即可。综上,我们需要一辅助数据结构如`pair`. ### Java ~~~ public class Solution { /** * @param nums: a list of integers * @return: find a majority number */ public int majorityNumber(ArrayList nums) { if (nums == null || nums.isEmpty()) return -1; // pair int key = -1, count = 0; for (int num : nums) { // re-initialize if (count == 0) { key = num; count = 1; continue; } // increment/decrement count if (key == num) { count++; } else { count--; } } return key; } } ~~~ ### 源码分析 初始化`count = 0`, 遍历数组时需要先判断`count == 0`以重新初始化。 ### 复杂度分析 时间复杂度 O(n)O(n)O(n), 空间复杂度 O(1)O(1)O(1).
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