Small sum problem and reverse pair problem

One, small and problem

In an array, the numbers on the left of each number that are smaller than the current number are added up, which is called the small sum of the array.

(1) Traverse the left O(N^2)

(2) Merge O(N)

Example: Find the small sum of [1, 3, 4, 2, 5]

1+（1+3）+1+（1+3+4+2）= 16

Code:

    public static int smallSum(int[] arr) {
		    if (arr == null || arr.length < 2) {
			return 0;
		}
		return mergeSort(arr, 0, arr.length - 1);
	}

	public static int mergeSort(int[] arr, int l, int r) {
		if (l == r) {
			return 0;
		}
		int mid = l + ((r - l) >> 1);
		return mergeSort(arr, l, mid) + mergeSort(arr, mid + 1, r) + merge(arr, l, mid, r);
	}

	public static int merge(int[] arr, int l, int m, int r) {
		int[] help = new int[r - l + 1];
		int i = 0;
		int p1 = l;
		int p2 = m + 1;
		int res = 0;
		while (p1 <= m && p2 <= r) {
			res += arr[p1] < arr[p2] ? (r - p2 + 1) * arr[p1] : 0;
			help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
		}
		while (p1 <= m) {
			help[i++] = arr[p1++];
		}
		while (p2 <= r) {
			help[i++] = arr[p2++];
		}
		for (i = 0; i < help.length; i++) {
			arr[l + i] = help[i];
		}
		return res;
	}

	// for test
	public static int comparator(int[] arr) {
		if (arr == null || arr.length < 2) {
			return 0;
		}
		int res = 0;
		for (int i = 1; i < arr.length; i++) {
			for (int j = 0; j < i; j++) {
				res += arr[j] < arr[i] ? arr[j] : 0;
			}
		}
		return res;
	}

	// for test
	public static int[] generateRandomArray(int maxSize, int maxValue) {
		int[] arr = new int[(int) ((maxSize + 1) * Math.random())];
		for (int i = 0; i < arr.length; i++) {
			arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random());
		}
		return arr;
	}

	// for test
	public static int[] copyArray(int[] arr) {
		if (arr == null) {
			return null;
		}
		int[] res = new int[arr.length];
		for (int i = 0; i < arr.length; i++) {
			res[i] = arr[i];
		}
		return res;
	}

	// for test
	public static boolean isEqual(int[] arr1, int[] arr2) {
		if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) {
			return false;
		}
		if (arr1 == null && arr2 == null) {
			return true;
		}
		if (arr1.length != arr2.length) {
			return false;
		}
		for (int i = 0; i < arr1.length; i++) {
			if (arr1[i] != arr2[i]) {
				return false;
			}
		}
		return true;
	}

	// for test
	public static void printArray(int[] arr) {
		if (arr == null) {
			return;
		}
		for (int i = 0; i < arr.length; i++) {
			System.out.print(arr[i] + " ");
		}
		System.out.println();
	}

 

Second, the reverse order problem

In an array, if the number on the left is greater than the number on the right, the two numbers form a reverse pair. The problem with Xiaohe is actually the same.