牛客刷题Java实现-----数组中的逆序对

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P。并将P对1000000007取模的结果输出。 即输出P%1000000007

第一种方法：暴力遍历

public class Solution {
	public static void main(String[] args) {
		
		//第一种方法：暴力遍历，复杂度 O（n^2）
		int[] array = {1,2,3,4,5,6,7,0};
		Solution s = new Solution();
		System.out.println(s.InversePairs(array));
	}
	public int InversePairs(int [] array) {
		int num = 0;
		for(int i=0;iarray[j]){
					num+=1;
				}
			}
		}
		int x = num%1000000007;
		return x;     
    }
}

第二种方法：归并实现

public class Solution {
	public static void main(String[] args) {
		int[] array = {1,2,3,4,5,6,7,0};
		Solution s = new Solution();
		System.out.println(s.InversePairs(array));
	}

	public int InversePairs(int[] array) {
		if (array == null || array.length <= 0) {
			return 0;
		}
		int pairsNum = mergeSort(array, 0, array.length - 1);
		return pairsNum;
	}

	public int mergeSort(int[] array, int left, int right) {
		if (left == right) {
			return 0;
		}
		int mid = (left + right) / 2;
		int leftNum = mergeSort(array, left, mid);
		int rightNum = mergeSort(array, mid + 1, right);
		return (Sort(array, left, mid, right) + leftNum + rightNum) % 1000000007;
	}

	public int Sort(int[] array, int left, int middle, int right) {
		int current1 = middle;
		int current2 = right;
		int current3 = right - left;
		int temp[] = new int[right - left + 1];
		int pairsnum = 0;
		while (current1 >= left && current2 >= middle + 1) {
			if (array[current1] > array[current2]) {
				temp[current3--] = array[current1--];
				pairsnum += (current2 - middle); // 这个地方是current2-middle！！！！
				if (pairsnum > 1000000007)// 数值过大求余
				{
					pairsnum %= 1000000007;
				}
			} else {
				temp[current3--] = array[current2--];
			}
		}
		while (current1 >= left) {
			temp[current3--] = array[current1--];
		}
		while (current2 >= middle + 1) {
			temp[current3--] = array[current2--];
		}
		// 将临时数组赋值给原数组
		int i = 0;
		while (left <= right) {
			array[left++] = temp[i++];
		}
		return pairsnum;
	}
}