分治-归并系列一＞数组中的逆序对

题目：

链接: link

解析：

代码：

class Solution {
    int[] tmp;
    public int reversePairs(int[] record) {
        tmp = new int[record.length];
        return mergeSort(record,0,record.length-1);
    }

    private int mergeSort(int[] record, int left, int right){
        if(left >= right) return 0;

        int ret = 0;
        int mid = (right + left) / 2;
        int cur1 = left, cur2 = mid+1, i = 0;

        //左半部分的个数 + 排序，右半部分的个数 + 排序
        ret += mergeSort(record,left, mid);
        ret += mergeSort(record, mid+1, right);

        //一左一右的个位数 + 排序
        while(cur1 <= mid && cur2 <= right){

            if(record[cur1] <= record[cur2]){
                tmp[i++] = record[cur1++];//排序
            } else {
                ret += mid - cur1 + 1;
                tmp[i++] = record[cur2++];//排序
            }
        } 

        while(cur1 <= mid) tmp[i++] = record[cur1++];
        while(cur2 <= right) tmp[i++] = record[cur2++];
        for(int j = left; j <= right; j++){
            record[j] = tmp[j-left];
        }

        return ret;
    }
}