劍指offer 51_數組中的逆序對(java)

面試題51. 數組中的逆序對

思路一 暴力法

會超出時間限制！！ 牛客中只能AC 50%

public class Solution {
    public int InversePairs(int [] array) {
        int count = 0;
        for(int i = 0; i < array.length; i++){
            for(int j = i+1; j < array.length; j++){
                if (array[i] > array[j]) count++;
            }
        }
        return count;
    }
}

思路一 暴力法

複雜度

• 時間： O(NlogN)
• 空間： O(N)

public class Solution {
    
    private int count = 0;
    public int InversePairs(int [] array) {
        if (array == null || array.length == 0) {
            return 0;
        }
        mergeSort(array);
        return count;
    }
    
    
    public void merge(int [] arr, int low, int high, int mid, int[] temp){
        if(low >= high) return ;
        int i = low;
        int j = mid+1;
        int p = low;
        // 合     ----  記錄左右兩邊的位置。
        while(i <= mid && j <= high){
            if(i <= mid && arr[i] <= arr[j]){
                temp[p++] = arr[i++];
            }else{
                temp[p++] = arr[j++];
                count = (count + (mid - i +1)) % 1000000007 ;
            }
        }
        while(i <= mid){
            temp[p++] = arr[i++];
        }
        while (j <= high){
            temp[p++] = arr[j++];
        }
        // 將temp中合併的結果拷貝到arr中
        p = low;
        while(low <= high){
            arr[low++] = temp[p++];
        }
    }
    // 分 -- 拆分爲左右兩個區間
    public void merge_sort(int[] arr, int low, int high, int[] temp){
        if(low < high){
            int mid = (low + high) / 2;
            merge_sort(arr, low, mid, temp);
            merge_sort(arr, mid+1, high, temp);
            merge(arr, low, high, mid, temp);
        }
    }
    
    public void mergeSort(int[] arr){
        int[] temp = new int[arr.length];
        merge_sort(arr, 0, arr.length-1, temp);
    }
}