leetcode 493. Reverse Pairs 逆序對數量 + 歸併排序做法
阿新 • • 發佈:2019-01-04
Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j].
You need to return the number of important reverse pairs in the given array.
Example1:
Input: [1,3,2,3,1]
Output: 2
Example2:
Input: [2,4,3,5,1]
Output: 3
Note:
The length of the given array will not exceed 50,000.
All the numbers in the input array are in the range of 32-bit integer.
題意很簡單,但是AC的方法不會,我覺得迴圈遍歷也不錯,雖然肯定會超時
程式碼如下:
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <cmath>
using namespace std;
class Solution
{
public:
int reversePairs(vector<int>& nums)
{
int res = sort_and_count(nums.begin(), nums.end());
return res;
}
int sort_and_count(vector<int>::iterator begin, vector <int>::iterator end)
{
if (end - begin <= 1)
return 0;
else
{
auto mid = begin + (end - begin) / 2;
int count = sort_and_count(begin, mid) + sort_and_count(mid, end);
for (auto i = begin, j = mid; i != mid; ++i)
{
while (j != end && (*i) > 2L * (*j) )
++j;
count += j - mid;
}
inplace_merge(begin, mid, end);
return count;
}
}
int reversePairsByLoop(vector<int>& nums)
{
int count = 0;
for (int i = 0; i < nums.size(); i++)
{
for (int j = i + 1; j < nums.size(); j++)
{
long long a = nums[i];
long long b = ((long long)nums[j]) << 1;
if (a > b)
count++;
}
}
return count;
}
};
#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#include <array>
using namespace std;
class Solution
{
public:
int reversePairs(vector<int>& a)
{
if (a.size() <= 0)
return 0;
vector<long long> all(a.begin(), a.end());
return mergeSort(all, 0, all.size() - 1);
}
long long mergeSort(vector<long long>& a, int begin, int end)
{
if (begin == end)
return 0;
else
{
long long mid = (end - begin) / 2 + begin;
long long leftCount = mergeSort(a, begin, mid);
long long rightCount = mergeSort(a, mid + 1, end);
long long midCount = 0;
long long i = begin, j = mid + 1;
while (i <= mid && j <= end)
{
if (a[i] <= 2 * a[j])
i++;
else
{
midCount += (mid - i + 1);
j++;
}
}
vector<long long> tmp;
i = begin, j = mid + 1;
while (i <= mid && j <= end)
{
if (a[i] <= a[j])
tmp.push_back(a[i++]);
else
tmp.push_back(a[j++]);
}
while (i <= mid)
tmp.push_back(a[i++]);
while (j <= end)
tmp.push_back(a[j++]);
for (int i = begin; i <= end; i++)
a[i] = tmp[i - begin];
return leftCount + midCount + rightCount;
}
}
};