leetcode Count of Smaller Numbers After Self

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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解题思路：水题一枚，从后向前处理，利用树状数组统计即可。其实应该先将数组中的数据离散化，之后用树状统计即可，无奈数据太水，随便搞搞就过了。

class Solution {public:    int tree[1000010];    void init() {        memset(tree, 0, sizeof(tree));    }    int lowbit(int x) {        return x&(-x);    }    void add(int x, int c) {        for(int i = x; i <= 1000000; i += lowbit(i)) {            tree[i] += c;        }        return ;    }    int getsum(int x) {        int ans = 0;        for(int i = x; i > 0; i -= lowbit(i)) {            ans += tree[i];        }        return ans;    }    vector<int> countSmaller(vector<int>& nums) {        vector<int> ans;        init();        int size = nums.size();        if(size == 0) return ans;        ans.push_back(0);        add(nums[size-1]+500000, 1);        for(int i = size - 2; i >= 0; --i) {            ans.push_back(getsum(nums[i]+500000-1));            add(nums[i]+500000, 1);        }        reverse(ans.begin(), ans.end());        return ans;    }};