[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。

 1 class Solution {

 2 public:

 3    /**

 4      * @param A: An integer array

 5      * @return: Count the number of element before this element 'ai' is 

 6      *          smaller than it and return count number array

 7      */

 8     int lowbit(int n) {

 9         return n & (-n);

10     }

11     

12     int sum(vector<int> &c, int n) {

13         int sum = 0;

14         while (n > 0) {

15             sum += c[n];

16             n -= lowbit(n);

17         }

18         return sum;

19     }

20     

21     void add(vector<int> &c, int i, int x) {

22         while (i < c.size()) {

23             c[i] += x;

24             i += lowbit(i);

25         }

26     }

27     

28     vector<int> countOfSmallerNumberII(vector<int> &A) {

29         // write your code here

30         vector<int> c(10002, 0);

31         vector<int> res(A.size());

32         for (int i = 0; i < A.size(); ++i) {

33             res[i] = sum(c, A[i]);

34             add(c, A[i] + 1, 1);

35         }

36         return res;

37     }

38 };

 

如果有负数或者数特别大的话，可以先离散化一下再搞。

 1 class Solution {

 2 public:

 3    /**

 4      * @param A: An integer array

 5      * @return: Count the number of element before this element 'ai' is 

 6      *          smaller than it and return count number array

 7      */

 8     int lowbit(int n) {

 9         return n & (-n);

10     }

11     

12     int sum(vector<int> &c, int n) {

13         int sum = 0;

14         while (n > 0) {

15             sum += c[n];

16             n -= lowbit(n);

17         }

18         return sum;

19     }

20     

21     void add(vector<int> &c, int i, int x) {

22         while (i < c.size()) {

23             c[i] += x;

24             i += lowbit(i);

25         }

26     }

27     

28     vector<int> countOfSmallerNumberII(vector<int> &A) {

29         // write your code here

30         vector<int> c(A.size() + 1, 0);

31         vector<int> res(A.size()), B(A);

32         map<int, int> mp;

33         sort(B.begin(), B.end());

34         for (int i = 0; i < B.size(); ++i) {

35             mp[B[i]] = i + 1;

36         }

37         for (int i = 0; i < A.size(); ++i) {

38             res[i] = sum(c, mp[A[i]] - 1);

39             add(c, mp[A[i]], 1);

40         }

41         return res;

42     }

43 };