《剑指Offer》之“数组中的逆序对”

题目描述

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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

代码实现

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c++版

class Solution {
    public:
    int allSum(vector<int> arr,int L,int mid,int R){
        int *help = new int(R-L+1);
        int i = 0 ;
        int pFirst = L;
        int pSecond = mid + 1;
        int sum = 0;
        while (pFirst <=mid && pSecond <=R) {
             sum += arr[pFirst] > arr[pSecond] ? (mid - pFirst+1):0;
              help[i++]  = arr[pFirst] < arr[pSecond] ? arr[pFirst++]:arr[pSecond++];
        }
        while (pFirst <= mid ) {
            help[i++] = arr[pFirst++];
        }
        while (pSecond <= R) {
            help[i++] = arr[pSecond++];
        }
        for (int k = 0; k < (R-L+1); k++) {
            arr[L+k] = help[k];
        }
        return sum;
    }
    int smallSum(vector<int> arr,int L,int R){
        if (L==R) {
            return 0;
        }
        int  mid = L+((R-L)>>1);
        //  相当于  int  mid = L+(R-L)/2;
        int leftSum = smallSum(arr, L, mid);
        int rightSum = smallSum(arr, mid + 1, R);
        int leftAndRightSum = allSum(arr,L,mid,R);
        return  leftSum + rightSum + leftAndRightSum;
        //    return smallSum(arr, L, mid)+smallSum(arr, mid + 1, R)+allSum(arr,L,mid,R);
    }
    int InversePairs(vector<int> data) {
        long length = data.size();
        if( length<2){
            return 0;
        }
        return smallSum(data,0,length );
    }
};

java版

public class Solution {
    public int InversePairs(int [] arr) {
        if (arr == null || arr.length < 2) {
            return 0;
        }
        return mergeSort(arr, 0, arr.length - 1)%1000000007;
    }

    public static int mergeSort(int[] arr, int l, int r) {
        if (l == r) {
            return 0;
        }
        int mid = (l + ((r - l) >> 1))%1000000007;
        return mergeSort(arr, l, mid)%1000000007 + mergeSort(arr, mid + 1, r) %1000000007+ merge(arr, l, mid, r)%1000000007;
    }
    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] ? (m - p1 + 1)%1000000007 : 0;
            res = res%1000000007;
            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;
    }
}