数组

归并排序

时间复杂度为 O(nlogn) 的排序算法之一，其中 n 是给定数组的长度。

///tc : O(nlogn)
//sc : O(n) for creating intermediate arrays a, b of size of part of subarray which is of size n
class Solution {
    public int[] sortArray(int[] nums) {
         merge(0,nums.length-1,nums);
         return nums;
    }
    public void merge(int start, int end, int arr[]){
        if(end>start){
            int mid = (start+end)/2;
            merge(start,mid,arr);
            merge(mid+1,end,arr);
            sort(start, mid,end, arr);
        }
    }
    public void sort(int start, int mid ,int end, int arr[]){
        int a[] = new int[mid-start+1];
        int b[] = new int[end-mid];
        for(int i = 0;i




<hr>

<p>反转计数</p>

<p>在数组排序之前需要进行多少次比较（给定数组 arr[] 的索引 i, j ，<strong>arr[i]> arr[j]</strong> （对于 j> i）将递增每次满足此条件，反转计数加 1。</p>

<p>注意：<em>我们可以使用相同的合并排序方法来查找反转计数（合并排序代码已稍微更改以使其更具可读性）</em><br>
</p>

<pre class="brush:php;toolbar:false">class Solution {
    // arr[]: Input Array
    // N : Size of the Array arr[]
    // Function to count inversions in the array.

    static long inversionCount(long arr[], int n) {
        // Your Code Here
       //we can use merge sort
       long temp[]= new long[n];
       return merge(0,n-1,arr,temp);


    }
    public static long merge(int start, int end, long arr[],long[] temp){
        long count = 0;
        if(end>start){
            int mid = (start+end)/2;
            count+=merge(start,mid,arr,temp);
            count+=merge(mid+1,end,arr,temp);
           count+=sort(start, mid,end, arr,temp);
        }
        return count;
    }
    public static long sort(int start, int mid ,int end, long arr[],long [] temp){
        long count = 0;
        int i = start;
        int j = mid+1;
        int k = start;

        while(i arr[j] then all the values after ith index including will be
                // greater that jth index value hence count += mid-i+1
            }
            k++;
        }
        while(i




          

            
  

            
        

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