归并排序
参考剑指 Offer 51. 数组中的逆序对 - 力扣（LeetCode） (leetcode-cn.com)
重点是被排序过程中的数字下标的对应，这里取了一个index数列，在排序的同时进行移动，从而保证下标对应
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| class Solution {
public:
vector<int> countSmaller(vector<int>& nums) {
int n = nums.size();
tmp.resize(n);
index.resize(n);
indextmp.resize(n);
ret.resize(n);
for(int i = 0; i < n; ++i)
index[i] = i;
mergeSort(nums, 0, n - 1);
return ret;
}
private:
vector<int> ret;
vector<int> tmp;
vector<int> index;
vector<int> indextmp;
void mergeSort(vector<int>& nums, int lo, int hi)
{
if(lo >= hi)
return;
int mid = lo + (hi - lo) / 2;
mergeSort(nums, lo, mid);
mergeSort(nums, mid + 1, hi);
copy(nums.begin() + lo, nums.begin() + hi + 1, tmp.begin() + lo);
copy(index.begin() + lo, index.begin() + hi + 1, indextmp.begin() + lo);
int i = lo, j = mid + 1, k = lo;
while(k <= hi)
{
if(i == mid + 1)
{
nums[k] = tmp[j];
index[k] = indextmp[j++];
}else if(j == hi + 1)
{
ret[indextmp[i]] += j - mid - 1;
nums[k] = tmp[i];
index[k] = indextmp[i++];
}
else if(tmp[i] <= tmp[j])
{
ret[indextmp[i]] += j - mid - 1;
nums[k] = tmp[i];
index[k] = indextmp[i++];
}else
{
nums[k] = tmp[j];
index[k] = indextmp[j++];
}
++k;
}
}
};
|
离散化树状数组
迟点看 https://www.cnblogs.com/xenny/p/9739600.html
https://leetcode-cn.com/problems/count-of-smaller-numbers-after-self/solution/shu-zhuang-shu-zu-by-liweiwei1419/