题目说明
给定一个整数数组 nums,按要求返回一个新数组 counts。数组 counts 有该性质: counts[i] 的值是 nums[i] 右侧小于 nums[i] 的元素的数量。
输入: [5,2,6,1]
输出: [2,1,1,0]
解释:
5 的右侧有 2 个更小的元素 (2 和 1).
2 的右侧仅有 1 个更小的元素 (1).
6 的右侧有 1 个更小的元素 (1).
1 的右侧有 0 个更小的元素.
链接:https://leetcode-cn.com/problems/count-of-smaller-numbers-after-self
分析:解题思路采用归并排序进行计算,借助c++中的pair结构
本题涉及的知识参考博客:
https://www.cnblogs.com/bigsai/p/12253254.html
https://www.cnblogs.com/hfut-freshguy/p/11565984.html
http://c.biancheng.net/view/6826.html
class Solution3
{
public:
vector<int> res;
vector<int> countSmaller(vector<int>& nums)
{
int n = nums.size();
res = vector<int>(n, 0);
if (n == 0)
return res;
vector<pair<int, int>> sortnums(n);
for (int i = 0; i < nums.size(); i++)
{
sortnums[i] = { nums[i], i };
}
mergeSort(sortnums, 0, n - 1);
return res;
}
void mergeSort(vector<pair<int, int>>& sortnums, int l, int r)
{
if (l < r)
{
int mid = (r - l) / 2 + l;
mergeSort(sortnums, l, mid);
mergeSort(sortnums, mid + 1, r);
merge(sortnums, l, mid, mid + 1, r);
}
}
void merge(vector<pair<int, int>>& sortnums, int begin1, int end1, int begin2, int end2)
{
int k = 0;
vector<pair<int, int>> tmp(end2 - begin1 + 1);
int i = begin1;
int j = begin2;
while (i <= end1 && j <= end2)
{
if (sortnums[i].first <= sortnums[j].first)
{
res[sortnums[i].second] += (j - end1 - 1);
tmp[k++] = sortnums[i++];
}
else
{
tmp[k++] = sortnums[j++];
}
}
while (i <= end1)
{
res[sortnums[i].second] += (j - end1 - 1);
tmp[k++] = sortnums[i++];
}
while (j <= end2)
{
tmp[k++] = sortnums[j++];
}
for (int i = begin1, j = 0; i <= end2; ++i, j++)
{
sortnums[i] = tmp[j];
}
}
};