Majority Element III

最新推荐文章于 2018-05-28 14:57:24 发布

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最新推荐文章于 2018-05-28 14:57:24 发布

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分类专栏： LintCode 题解

本文链接：https://blog.youkuaiyun.com/github_34333284/article/details/51695456

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本文介绍了一种寻找数组中多数元素的方法，该元素出现次数超过数组长度的1/k。使用大小为k的哈希映射来记录可能的多数元素，并通过两遍数组扫描确保找到正确答案。

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#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

/*
  Given an array of integers and a number K, the majority number is the number
  that occurs more than 1/k if the size of the array.
  Note: there is only one majoirty number in the array.
*/

/*
  if we use counting method, it will take too much memory.
  Since we only need to get the only one majority number. we can use a size K hashmap
  to remember the current majority.
*/

int majorityNumber(vector<int>& nums, int k) {
  unordered_map<int, int> numToCount;
  int n = nums.size();
  for(const auto& i : nums) {
    ++numToCount[i];
    if(numToCount.size() == k) {
      auto iter = numToCount.begin();
      while(iter != numToCount.end()) {
        if(--(iter->second) == 0) {
          numToCount.erase(iter++);
        } else ++iter;
      }
    }
  }
  for(auto& it : numToCount) {
    it.second = 0;
  }
  vector<int> res;
  for(const auto& i : nums) {
    auto it = numToCount.find(i);
    if(it.second > static_cast<double>(n) / k) {
      res.push_back(it.first);
    }
  }
  return res[0];
}