[LintCode] Count of Smaller Number before itself

Count of Smaller Number before itself

Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) . For each element Ai in the array, count the number of element before this element Ai is smaller than it and return count number array.

Example

For array [1,2,7,8,5], return [0,1,2,3,2]

Note

We suggest you finish problem Segment Tree Build, Segment Tree Query II and Count of Smaller Number before itself I first.

题目让用线段树，其实树状数组就能搞定，而且树状数组的代码太短小精悍了。

class Solution {
public:
   /**
     * @param A: An integer array
     * @return: Count the number of element before this element 'ai' is 
     *          smaller than it and return count number array
     */
    int lowbit(int n) {
        return n & (-n);
    }
    
    int sum(vector<int> &c, int n) {
        int sum = 0;
        while (n > 0) {
            sum += c[n];
            n -= lowbit(n);
        }
        return sum;
    }
    
    void add(vector<int> &c, int i, int x) {
        while (i < c.size()) {
            c[i] += x;
            i += lowbit(i);
        }
    }
    
    vector<int> countOfSmallerNumberII(vector<int> &A) {
        // write your code here
        vector<int> c(10002, 0);
        vector<int> res(A.size());
        for (int i = 0; i < A.size(); ++i) {
            res[i] = sum(c, A[i]);
            add(c, A[i] + 1, 1);
        }
        return res;
    }
};

如果有负数或者数特别大的话，可以先离散化一下再搞。

class Solution {
public:
   /**
     * @param A: An integer array
     * @return: Count the number of element before this element 'ai' is 
     *          smaller than it and return count number array
     */
    int lowbit(int n) {
        return n & (-n);
    }
    
    int sum(vector<int> &c, int n) {
        int sum = 0;
        while (n > 0) {
            sum += c[n];
            n -= lowbit(n);
        }
        return sum;
    }
    
    void add(vector<int> &c, int i, int x) {
        while (i < c.size()) {
            c[i] += x;
            i += lowbit(i);
        }
    }
    
    vector<int> countOfSmallerNumberII(vector<int> &A) {
        // write your code here
        vector<int> c(A.size() + 1, 0);
        vector<int> res(A.size()), B(A);
        map<int, int> mp;
        sort(B.begin(), B.end());
        for (int i = 0; i < B.size(); ++i) {
            mp[B[i]] = i + 1;
        }
        for (int i = 0; i < A.size(); ++i) {
            res[i] = sum(c, mp[A[i]] - 1);
            add(c, mp[A[i]], 1);
        }
        return res;
    }
};