牛客20、5（归并排序）

20数组中的逆序对

 

 

代码

public class Solution {
    int count = 0;
    public int InversePairs(int [] array) {
        if(array.length < 2)
            return 0;
        int[] result = new int[array.length];
        mergeSort(array, 0, array.length - 1, result);
        return count;
    }
    public void mergeSort(int[] array, int left, int right, int[] result) {
        if (left == right) return;
        int mid = left + (right - left) / 2;

        mergeSort(array, left, mid, result);
        mergeSort(array, mid + 1, right, result);
        merge(array, left, right, result);
        
    }
    public void merge(int[] array, int start, int end, int[] result) {
        int end1 = (start + end) / 2;
        int start2 = end1 + 1;
        int index1 = start;
        int index2 = start2;
        while (index1 <= end1 && index2 <= end) {
            if (array[index1] <= array[index2])
                result[index1 + index2 - start2] = array[index1++];
            else {
                result[index1 + index2 - start2] = array[index2++];
                count += start2 - index1;
                count %= 1000000007;
            }
        }
        while (index1 <= end1) 
            result[index1 + index2 - start2] = array[index1++];
        while (index2 <= end)
            result[index1 + index2 - start2] = array[index2++];
        while (start <= end)
            array[start] = result[start++];
    }
}

 

 

 

5合并k个已排序的链表

public class Solution {
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        return divideMerge(lists, 0, lists.size() - 1);
    }
    public ListNode  divideMerge(ArrayList<ListNode> lists, int left, int right) {
        if (left > right) {
            return null;
        } else if (left == right) {
            return lists.get(left);
        }
        int mid = (left + right) / 2;
        return Merge(divideMerge(lists, left, mid), divideMerge(lists, mid + 1, right));
    }
    public ListNode Merge(ListNode list1, ListNode list2) {
        if (list1 == null) return list2;
        if (list2 == null) return list1;
        ListNode head = new ListNode(0);
        ListNode cur = head;
        while (list1 != null && list2 != null) {
            if (list1.val <= list2.val) {
                cur.next = list1;
                list1 = list1.next;
            } else {
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
            if (list1 != null) {
                cur.next = list1;
            } else {
                cur.next = list2;
            }
        }
        return head.next;
    }
}