feat: add python and java solutions to lcof question

添加《剑指 Offer》题解：面试题51. 数组中的逆序对

.github/workflows/sync.yml

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+name: Sync
+
+on:
+  push:
+    branches: [ master ]
+
+jobs:
+  build:
+    runs-on: ubuntu-latest
+    steps:
+    - name: Sync to Gitee
+      uses: wearerequired/git-mirror-action@master
+      env:
+          SSH_PRIVATE_KEY: ${{ secrets.GITEE_RSA_PRIVATE_KEY }}
+      with:
+          source-repo: "[email protected]:doocs/leetcode.git"
+          destination-repo: "[email protected]:Doocs/leetcode.git"

lcof/面试题51. 数组中的逆序对/README.md

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+# [面试题51. 数组中的逆序对](https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/)
+
+## 题目描述
+<!-- 这里写题目描述 -->
+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。
+
+**示例 1:**
+
+```
+输入: [7,5,6,4]
+输出: 5
+```
+
+**限制：**
+
+- `0 <= 数组长度 <= 50000`
+
+## 解法
+<!-- 这里可写通用的实现逻辑 -->
+在归并中统计逆序对。
+
+### Python3
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def reversePairs(self, nums: List[int]) -> int:
+        self.res = 0
+
+        def merge(part1, part2, nums):
+            len1, len2 = len(part1) - 1, len(part2) - 1
+            t = len(nums) - 1
+            while len1 >= 0 and len2 >= 0:
+                if part1[len1] > part2[len2]:
+                    self.res += (len2 + 1)
+                    nums[t] = part1[len1]
+                    len1 -= 1
+                else:
+                    nums[t] = part2[len2]
+                    len2 -= 1
+                t -= 1
+            while len1 >= 0:
+                nums[t] = part1[len1]
+                t -= 1
+                len1 -= 1
+            while len2 >= 0:
+                nums[t] = part2[len2]
+                t -= 1
+                len2 -= 1
+
+        def merge_sort(nums):
+            if len(nums) < 2:
+                return
+            mid = len(nums) // 2
+            s1, s2 = nums[:mid], nums[mid:]
+            merge_sort(s1)
+            merge_sort(s2)
+            merge(s1, s2, nums)
+
+        merge_sort(nums)
+        return self.res
+
+```
+
+### Java
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    private int res = 0;
+    public int reversePairs(int[] nums) {
+        int n = nums.length;
+        if (n < 2) {
+            return 0;
+        }
+        mergeSort(nums, 0, n - 1);
+        return res;
+    }
+
+    private void mergeSort(int[] nums, int s, int e) {
+        if (s == e) {
+            return;
+        }
+        int mid = s + ((e - s) >> 1);
+        mergeSort(nums, s, mid);
+        mergeSort(nums, mid + 1, e);
+        merge(nums, s, mid, e);
+    }
+
+    private void merge(int[] nums, int s, int mid, int e) {
+        int n = e - s + 1;
+        int[] help = new int[n];
+        int i = s, j = mid + 1, idx = 0;
+        while (i <= mid && j <= e) {
+            if (nums[i] > nums[j]) {
+                res += (mid - i + 1);
+                help[idx++] = nums[j++];
+            } else {
+                help[idx++] = nums[i++];
+            }
+        }
+        while (i <= mid) {
+            help[idx++] = nums[i++];
+        }
+        while (j <= e) {
+            help[idx++] = nums[j++];
+        }
+        for (int t = 0; t < n; ++t) {
+            nums[s + t] = help[t];
+        }
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题51. 数组中的逆序对/Solution.java

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+class Solution {
+    private int res = 0;
+    public int reversePairs(int[] nums) {
+        int n = nums.length;
+        if (n < 2) {
+            return 0;
+        }
+        mergeSort(nums, 0, n - 1);
+        return res;
+    }
+
+    private void mergeSort(int[] nums, int s, int e) {
+        if (s == e) {
+            return;
+        }
+        int mid = s + ((e - s) >> 1);
+        mergeSort(nums, s, mid);
+        mergeSort(nums, mid + 1, e);
+        merge(nums, s, mid, e);
+    }
+
+    private void merge(int[] nums, int s, int mid, int e) {
+        int n = e - s + 1;
+        int[] help = new int[n];
+        int i = s, j = mid + 1, idx = 0;
+        while (i <= mid && j <= e) {
+            if (nums[i] > nums[j]) {
+                res += (mid - i + 1);
+                help[idx++] = nums[j++];
+            } else {
+                help[idx++] = nums[i++];
+            }
+        }
+        while (i <= mid) {
+            help[idx++] = nums[i++];
+        }
+        while (j <= e) {
+            help[idx++] = nums[j++];
+        }
+        for (int t = 0; t < n; ++t) {
+            nums[s + t] = help[t];
+        }
+    }
+}

lcof/面试题51. 数组中的逆序对/Solution.py

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+class Solution:
+    def reversePairs(self, nums: List[int]) -> int:
+        self.res = 0
+
+        def merge(part1, part2, nums):
+            len1, len2 = len(part1) - 1, len(part2) - 1
+            t = len(nums) - 1
+            while len1 >= 0 and len2 >= 0:
+                if part1[len1] > part2[len2]:
+                    self.res += (len2 + 1)
+                    nums[t] = part1[len1]
+                    len1 -= 1
+                else:
+                    nums[t] = part2[len2]
+                    len2 -= 1
+                t -= 1
+            while len1 >= 0:
+                nums[t] = part1[len1]
+                t -= 1
+                len1 -= 1
+            while len2 >= 0:
+                nums[t] = part2[len2]
+                t -= 1
+                len2 -= 1
+
+        def merge_sort(nums):
+            if len(nums) < 2:
+                return
+            mid = len(nums) // 2
+            s1, s2 = nums[:mid], nums[mid:]
+            merge_sort(s1)
+            merge_sort(s2)
+            merge(s1, s2, nums)
+
+        merge_sort(nums)
+        return self.res