剑指 Offer 51. 数组中的逆序对（困难）_sword point to offer51. reverse pairs in an array-优快云博客

本文链接：https://blog.youkuaiyun.com/aninstein/article/details/117634023

该篇博客详细介绍了如何利用归并排序和离散化树状数组两种方法解决《剑指Offer》中数组中的逆序对问题。作者提供了完整的Java代码实现，并附带了官方题解链接，帮助读者理解这两种高效算法的思路和应用。

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剑指 Offer 51. 数组中的逆序对

题目链接

https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/

代码连接：

https://gitee.com/aninstein/HappyJava/blob/master/learn_java/src/leetcode/offer100/arrays/HardReversePairs.java

题目解析：

方法1：归并排序
方法2：离散化树状数组

官方题解：https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/solution/shu-zu-zhong-de-ni-xu-dui-by-leetcode-solution/

package leetcode.offer100.arrays;


import base.Utils;

/**
 * 题目：剑指 Offer 51. 数组中的逆序对
 * 题目连接：https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/
 *
 * 方法1：归并排序
 * 方法2：离散化树状数组
 * 官方题解：https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/solution/shu-zu-zhong-de-ni-xu-dui-by-leetcode-solution/
 *
 */
public class HardReversePairs {

    /**
     * 1. 使用归并排序解题
     * @param nums
     * @return
     */
    public static int reversePairs(int[] nums) {
        int len = nums.length;

        if (len < 2) {
            return 0;
        }
        int[] temp = new int[len];
        return countReversePairs(nums, temp, 0, len - 1);
    }

    private static int countReversePairs(int[] nums, int[] temp, int left, int right) {
        if (left == right) {
            return 0;
        }

        int mid = Utils.midNumber(left, right);
        int leftPairs = countReversePairs(nums, temp, left, mid);
        int rightPairs = countReversePairs(nums, temp, mid + 1, right);

        if (nums[mid] <= nums[mid + 1]) {
            return leftPairs + rightPairs;
        }

        int crossPairs = countAndMerge(nums, temp, left, mid, right);
        return leftPairs + crossPairs + rightPairs;
    }

    private static int countAndMerge(int[] nums, int[] temp, int left, int mid, int right) {
        for (int i = left; i <= right ; i++) {
            temp[i] = nums[i];
        }

        int i = left, j = mid + 1;
        int count = 0;
        for (int k = left; k <= right; k++) {
            if (i == mid + 1) {
                nums[k] = temp[j++];
            } else if (j == right + 1) {
                nums[k] = temp[i++];
            } else if (temp[i] <= temp[j]) {
                nums[k] = temp[i++];
            } else {
                nums[k] = temp[j++];
                count += (mid - i + 1);
            }
        }
        return count;
    }

    public static void main(String[] args) {

    }

}