题目：
在数组中的两个数字如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组，求出这个数组中的逆序对的总数。
解法：
暴力遍历的方法，直观可行，O(n^2)
基于归并排序的解法：
int inversePairs(int* data, int length) {
if (data == 0 || length <= 0) return 0;
int *copy = new int[length];
for (int i = 0; i < length; ++i) {
copy[i] = data[i];
}
int cnt = inversePairsCore(data, copy, 0, length - 1);
delete[] copy;
return cnt;
}
int inversePairsCore(int* data, int* copy, int start, int end) {
if (start == end) {
copy[start] = data[start];
return 0;
}
int length = (end - start)/2;
int left = inversePairsCore(data, copy, start, start + length);
int right = inversePairsCore(data, copy, start + length + 1, end);
int i = start + length;
int j = end;
int indexCopy = end;
int cnt = 0;
while (i >= start && j >= start+length+1) {
if (data[i] > data[j]) {
copy[indexCopy--] = data[i--];
cnt += j - start - length;
} else {
copy[indexCopy--] = data[j--];
}
}
for (; i >= start; --i) {
copy[indexCopy--] = data[i--];
}
for (; j >= start+length+1; --j) {
copy[indexCopy--] = data[j--];
}
return left + right + cnt;
}
