Leetcode 315. Count of Smaller Numbers After Self[hard]

题目：
You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Given nums = [5, 2, 6, 1]

To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
Return the array [2, 1, 1, 0].

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这个题和Leetcode 327. Count of Range Sum[hard]（http://blog.csdn.net/qq379548839/article/details/52601135）很像，也可以用相同的做法：再平衡树中倒叙插入数组中的每个数字，在插入时可以计算出之前插入的小于这个数字的数字个数，即为counts。同样的，也可以用离散化+树状数组。

但有一种新的方法：分治。

将数组分成左右两个部分，则左边部分一个数字lx对应的答案=右边部分比lx小的数字的个数+左边部分内部lx右边比lx小的数字个数；右边部分一个数字rx对应的答案=右边部分内部rx右边比rx小的数字个数。
而这个程类可以在归并排序过程中轻易实现。

同样的，这种分治方法同样可以用在Leetcode 327. Count of Range Sum[hard]中~

注意：这个题有个坑，传入的nums可能为空。答案需要特判一下。

class Solution {public:    vector<int> v;    vector<int> cnt;    vector<int> s;    vector<int> countSmaller(vector<int>& nums) {        v.clear();        cnt.clear();        if (nums.size() == 0) return cnt;        for (int i = 0; i < nums.size(); i++) cnt.push_back(0);        for (int i = 0; i < nums.size(); i++) v.push_back(i);        calc(nums, 0, nums.size() - 1);        return cnt;    }    void calc(vector<int>& nums, int lt, int rt) {        if (lt == rt) return;        int x = (lt + rt) / 2;        calc(nums, lt, x);        calc(nums, x + 1, rt);        s.clear();        int pa = lt, pb = x + 1;        while (pa <= x || pb <= rt) {            if (pa > x) {                s.push_back(v[pb]);                pb++;                //if (lt == 0 && rt == 3) cout << "a" << endl;            }            else if (pb > rt) {                s.push_back(v[pa]);                cnt[v[pa]] += pb - (x + 1);                pa++;                //if (lt == 0 && rt == 3) cout << "b" << endl;            }            else {                if (nums[v[pa]] <= nums[v[pb]]) {                    s.push_back(v[pa]);                    cnt[v[pa]] += pb - (x + 1);                    pa++;                    //if (lt == 0 && rt == 3) cout << "c" << endl;                }                else {                    s.push_back(v[pb]);                    pb++;                    //if (lt == 0 && rt == 3) cout << "d" << endl;                }            }        }        for (int i = 0; i < s.size(); i++) v[i + lt] = s[i];        return;    }};