2019-11-02 12:07:41 +08:00
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# 51. 数组中的逆序对
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[NowCoder](https://www.nowcoder.com/practice/96bd6684e04a44eb80e6a68efc0ec6c5?tpId=13&tqId=11188&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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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。
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## 解题思路
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```java
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private long cnt = 0;
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private int[] tmp; // 在这里声明辅助数组，而不是在 merge() 递归函数中声明
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public int InversePairs(int[] nums) {
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tmp = new int[nums.length];
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mergeSort(nums, 0, nums.length - 1);
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return (int) (cnt % 1000000007);
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}
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private void mergeSort(int[] nums, int l, int h) {
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if (h - l < 1)
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return;
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int m = l + (h - l) / 2;
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mergeSort(nums, l, m);
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mergeSort(nums, m + 1, h);
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merge(nums, l, m, h);
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}
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private void merge(int[] nums, int l, int m, int h) {
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int i = l, j = m + 1, k = l;
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while (i <= m || j <= h) {
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if (i > m)
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tmp[k] = nums[j++];
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else if (j > h)
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tmp[k] = nums[i++];
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else if (nums[i] <= nums[j])
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tmp[k] = nums[i++];
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else {
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tmp[k] = nums[j++];
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this.cnt += m - i + 1; // nums[i] > nums[j]，说明 nums[i...mid] 都大于 nums[j]
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}
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k++;
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}
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for (k = l; k <= h; k++)
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nums[k] = tmp[k];
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}
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```
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2019-11-02 14:39:13 +08:00
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2019-11-02 17:33:10 +08:00
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
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