51-数组中的逆序对

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。

思路使用归并排序
首先要会写归并排序：

public class MergeSort {
    private static Comparable[] aux;

    public static void sort(Comparable[] a) {
        aux = new Comparable[a.length];
        sort(a, 0, a.length - 1);
    }

    private static void sort(Comparable[] a, int lo, int hi) {
        if (lo >= hi) return;
        int mid = lo + (hi - lo) / 2;
        sort(a, lo, mid);
        sort(a, mid + 1, hi);
        merge(a, lo, mid, hi);
    }

    private static void merge(Comparable[] a, int lo, int mid, int hi) {
        int i = lo, j = mid + 1;
        for (int k = lo; k <= hi; k++) aux[k] = a[k];
        for (int k = lo; k <= hi; k++) {
            if (i > mid) {
                a[k] = aux[j++];
            } else if (j > hi) {
                a[k] = aux[i++];
                // 必须是 less(aux[j], aux[i])，不能为less(aux[i], aux[j])
                // 保证归并排序稳定性
            } else if (less(aux[j], aux[i])) {
                a[k] = aux[j++];
            } else {
                a[k] = aux[i++];
            }
        }
    }

    private static boolean less(Comparable a, Comparable b) {
        return a.compareTo(b) < 0;
    }
}

答案：

class Solution {
    private int[] aux;
    public int reversePairs(int[] a) {
        if (a == null || a.length == 0) return 0;
        aux = new int[a.length];
        return getCount(a, 0, a.length - 1);
    }

    private int getCount(int[] a, int lo, int hi) {
        if (lo >= hi) return 0;
        int mid = lo + (hi - lo) / 2;
        int left = getCount(a, lo, mid);
        int right = getCount(a, mid + 1, hi);
        if (a[mid] <= a[mid + 1]) {
            return left + right;
        }
        int cross = merge(a, lo, mid, hi);
        return left + right + cross;
    }

    private int merge(int[] a, int lo, int mid, int hi) {
        int i = lo, j = mid + 1, count = 0;
        for (int k = lo; k <= hi; k++) aux[k] = a[k];
        for (int k = lo; k <= hi; k++) {
            if (i > mid) {
                a[k] = aux[j++];
            } else if (j > hi) {
                a[k] = aux[i++];
            } else if (aux[j] < aux[i]) {
                a[k] = aux[j++];
                count += mid - i + 1;
            } else {
                a[k] = aux[i++];
            }
        }
        return count;
    }
}