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前言

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就这么说吧，所有递归的算法，你甭管它是干什么的，本质上都是在遍历一棵（递归）树，然后在节点（前中后序位置）上执行代码，你要写递归算法，本质上就是要告诉每个节点需要做什么

一、力扣912. 排序数组

class Solution {public int[] sortArray(int[] nums) {Merge.sort(nums);return nums;}
}
class Merge{public static int[] temp;public static void sort(int[] nums){temp = new int[nums.length];sort(nums, 0, nums.length-1);}public static void sort(int[] nums, int low, int high){if(low == high){return;}int mid = low + (high-low)/2;sort(nums, low, mid);sort(nums,mid+1,high);merge(nums,low,high,mid);}public static void merge(int[] nums, int low, int high, int mid){for(int a = low; a <= high; a ++){temp[a] = nums[a];}int i = low, j = mid+1, k = low;for(; i <= mid && j <= high;){if(temp[i] < temp[j]){nums[k++] = temp[i++];}else{nums[k++] = temp[j++];}}while(i <= mid){nums[k++] = temp[i++];}while(j <= high){nums[k++] = temp[j++];}}
}

二、力扣 315. 计算右侧小于当前元素的个数

class Solution {private class Pair{int id;int val;public Pair(int id, int val){this.id = id;this.val = val;}}Pair[] temp;int[] count;public List<Integer> countSmaller(int[] nums) {int n = nums.length;temp = new Pair[n];Pair[] arr = new Pair[n];for(int i = 0; i < n; i ++){arr[i] = new Pair(i,nums[i]);}count = new int[n];sort(arr,0,n-1);List<Integer> res = new ArrayList<>();for(int i = 0; i < n; i ++){res.add(count[i]);}return res;}public void sort(Pair[] arr, int low, int high){if(low == high){return;}int mid = low + (high-low)/2;sort(arr,low,mid);sort(arr,mid+1,high);merge(arr,low,high,mid);}public void merge(Pair[] arr, int low, int high, int mid){for(int i = low; i <= high; i ++){temp[i] = arr[i];}for(int i = low, j = mid+1, k = low; k <= high; k ++){if(i == mid+1){arr[k] = temp[j++];}else if(j == high+1){count[temp[i].id] += j-mid-1;arr[k] = temp[i++];}else if(temp[i].val > temp[j].val){arr[k] = temp[j++];}else{count[temp[i].id] += j-mid-1;arr[k] = temp[i++];}}}
}

三、力扣493. 翻转对

class Solution {int[] temp;int count = 0;public int reversePairs(int[] nums) {temp = new int[nums.length];sort(nums,0, nums.length-1);return count;}public void sort(int[] nums, int low, int high){if(low == high){return;}int mid = low + (high-low)/2;sort(nums,low,mid);sort(nums,mid+1,high);merge(nums,low,high,mid);}public void merge(int[] nums, int low, int high, int mid){for(int i = low; i <= high; i ++){temp[i] = nums[i];}int end = mid+1;for(int i = low; i <= mid; i ++){while(end <= high && (long)nums[i] > (long)2*nums[end]){end ++;}count += end - mid -1;}for(int i = low, j = mid+1, k = low; k <= high; k ++){if(i == mid+1){nums[k] = temp[j++];}else if(j == high+1){nums[k] = temp[i++];}else if(temp[i] > temp[j]){   nums[k] = temp[j++];}else{nums[k] = temp[i++];}}}
}

四、力扣327. 区间和的个数

class Solution {int lower,upper,count=0;long[] temp;public int countRangeSum(int[] nums, int lower, int upper) {this.lower= lower;this.upper = upper;long[] preSum = new long[nums.length+1];for (int i = 0; i < nums.length; i++) {preSum[i + 1] = (long)nums[i] + preSum[i];}temp = new long[preSum.length];sort(preSum, 0, preSum.length-1);return count;}public void sort(long[] nums, int low, int high){if(low == high){return;}int mid = low + (high-low)/2;sort(nums,low,mid);sort(nums,mid+1,high);merge(nums,low,high,mid);}public void merge(long[] nums,int low, int high, int mid){for(int i = low; i <= high; i ++){temp[i] = nums[i];}int start = mid+1, end = mid+1;for(int i = low; i <= mid; i ++){while(start <= high && nums[start]-nums[i] < lower){start ++;}while(end <= high && nums[end] - nums[i] <= upper){end ++;}count += end-start;}for(int i = low, j = mid+1, k = low; k <= high; k ++){if(i == mid+1){nums[k] = temp[j++];}else if(j == high+1){nums[k] = temp[i++];}else if(temp[i] > temp[j]){nums[k] = temp[j++];}else{nums[k] = temp[i++];}}}
}