[剑指Offer][数组]数组中的逆序对

题目描述

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。

 

public class Solution {
    public int InversePairs(int[] array) {
        if(array.length == 0) {
            return 0;
        }
        int result = 0;
        for(int i = 0; i < array.length - 1; i ++) {
            for(int j = i + 1; j < array.length; j ++) {
                if(array[i] > array[j]) {
                    result ++;
                }
            }
        }
        return result % 1000000007;
    }
}

public class Solution {
    public int InversePairs(int[] array) {
        return mergeSort(array, 0, array.length - 1);
    }

    public int mergeSort(int[] data, int left, int right) {
        if(left >= right) {
            return 0;
        }
        int inversePairNum = 0;
        int middle = (left + right) / 2;
        inversePairNum = (inversePairNum + mergeSort(data, left, middle)) % 1000000007;
        inversePairNum = (inversePairNum + mergeSort(data, middle + 1, right)) % 1000000007;
        inversePairNum = (inversePairNum + mergeArray(data, left, middle, right)) % 1000000007;
        return inversePairNum;
    }

    public int mergeArray(int[] data, int left, int middle, int right) {
        int[] temp = new int[right - left + 1];
        int pointerOne = left;
        int pointerTwo = middle + 1;
        int count = 0;
        int inversePairNum = 0;
        for(count = 0; count < temp.length; count ++) {
            if((pointerOne > middle) || (pointerTwo > right)) {
                break;
            }
            if(data[pointerOne] <= data[pointerTwo]) {
                temp[count] = data[pointerOne ++];
            }
            else {
                temp[count] = data[pointerTwo ++];
                inversePairNum = (inversePairNum + (middle - pointerOne + 1)) % 1000000007;
            }
        }
        while(pointerOne <= middle) {
            temp[count ++] = data[pointerOne ++];
        }
        while(pointerTwo <= right) {
            temp[count ++] = data[pointerTwo ++];
        }
        for(int i = 0; i < count; i ++) {
            data[left + i] = temp[i];
        }
        return inversePairNum;
    }
}