LintCode 532: Reverse Pairs (Fenwick tree, Merge Sort classic title)

532. Reverse Pairs
     中文English
     Reverse pair is a pair of numbers (A[i], A[j]) such that A[i] > A[j] and i < j. Given an array, return the number of reverse pairs in the array.

Example
Example1

Input: A = [2, 4, 1, 3, 5]
Output: 3
Explanation:
(2, 1), (4, 1), (4, 3) are reverse pairs
Example2

Input: A = [1, 2, 3, 4]
Output: 0
Explanation:
No reverse pair

Solution 1: Fenwick tree
reference online answer.
note:

1. The solution of the C [x] x represents the inside of the actual value of x, instead of the subscript.
2. Code words discrete useful [3,2,100000] may be simplified to [2,1,3], and does not affect the results. Note the last to +1.

    for (int i = 0; i < A.size(); ++i) {
            A[i] = lower_bound(sortedA.begin(), sortedA.begin() + uniqLen, A[i]) - sortedA.begin() + 1;
    }

3. Note to heavy, so use the unique () function is very useful.
4. A [i] to be from the beginning, C [] A ratio of size [] over.
5. Two line sequential calculation result for loop can be interchanged.

            result += sum(uniqLen) - sum(A[i]);
            add(A[i], 1);

code show as below:

class Solution {
public:
    /**
     * @param A: an array
     * @return: total of reverse pairs
     */
    long long reversePairs(vector<int> &A) {
        vector<int> sortedA = A;
        sort(sortedA.begin(), sortedA.end());
        uniqLen = unique(sortedA.begin(), sortedA.end()) - sortedA.begin();
        C.resize(uniqLen + 1, 0);
        long long result = 0;

        //discrete A[]
        for (int i = 0; i < A.size(); ++i) {
            A[i] = lower_bound(sortedA.begin(), sortedA.begin() + uniqLen, A[i]) - sortedA.begin() + 1;
        }
        
        for (int i = 0; i < A.size(); ++i) {
            result += sum(uniqLen) - sum(A[i]);
            add(A[i], 1);
        }
        
        return result;
    }


private:
    int uniqLen;
    vector<int> C;
    
    int lowbit(int x) {
        return x & (-x);
    }
    
    int sum(int x) {
        int result = 0;
        for (int i = x; i > 0; i -= lowbit(i)) {
            result += C[i];
        }
        return result;
    }
    
    void add(int x, int v) {
        for (int i = x; i <= uniqLen; i += lowbit(i)) {
            C[i] += v;
        }
    }
};

Method 2: Similar to Method 1, but similarly constructed with C brutal force []. Time complexity of O (n ^ 2).
Time out.

class Solution {
public:
    /**
     * @param A: an array
     * @return: total of reverse pairs
     */
    long long reversePairs(vector<int> &A) {
        vector<int> sortedA = A;
        sort(sortedA.begin(), sortedA.end());
        uniqLen = unique(sortedA.begin(), sortedA.end()) - sortedA.begin();
        C.resize(uniqLen + 1, 0);
        long long result = 0;

        //discrete A[]
        for (int i = 0; i < A.size(); ++i) {
            A[i] = lower_bound(sortedA.begin(), sortedA.begin() + uniqLen, A[i]) - sortedA.begin() + 1;
        }
        
        for (int i = 0; i < A.size(); ++i) {
            for (int j = A[i]; j <= uniqLen; ++j) {
                C[j]++;
            }
            result += C[uniqLen] - C[A[i]];
        }
        
        return result;
    }

private:
    int uniqLen;
    vector<int> C;
};

Solution 3: Merge Sort
need to add this line:
Result = MID + -. 1 left +;

class Solution {
public:
    /**
     * @param A: an array
     * @return: total of reverse pairs
     */
    long long reversePairs(vector<int> &A) {
        buf.resize(A.size(), 0);
        return mergeSort(A, buf, 0, A.size() - 1);
    }
private:
    long long mergeSort(vector<int> & A, vector<int> & buf, int start, int end) {
        
        if (start >= end) return 0;
        
        int result = 0;
        int mid = start + (end - start) / 2;
        
        result += mergeSort(A, buf, start, mid);
        result += mergeSort(A, buf, mid + 1, end);
        result += merge(A, buf, start, end);
        
        return result;
    }
    
    long long merge(vector<int> & A, vector<int> & buf, int start, int end) {
        int result = 0;
        int mid = start + (end - start) / 2;
        int left = start, right = mid + 1, index = start;
        
        while(left <= mid && right <= end) {
            if (A[left] <= A[right]) {
                buf[index++] = A[left++];
            } else {
                buf[index++] = A[right++];
                result += mid - left + 1;
            }
        }
        
        while(left <= mid) {
            buf[index++] = A[left++];
        }
        
        while(right <= end) {
            buf[index++] = A[right++];
        }
        
        for (index = start; index <= end; index++) {
            A[index] = buf[index];
        }
        
        return result;
    }
    
    vector<int> buf;
};