面试题 36: 数组中的逆序对

一. 题目

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.

代码请到我的代码库中下载 Point2Offer

二. 代码

package ween_2;/**难度系数:**** * 剑指offer: 数组中的逆序对 * 方法:归并排序思想 * 测试用例:功能测试(123456,135246,654321,52135,31,1) * 特殊输入测试(数组为空) * @author dingding * Date:2017-6-22 7:30 * Declaration: All Rights Reserved! */public class No36 {    public static void main(String[] args) {        test1();        test2();        test3();        test4();        test5();        test6();        test7();    }    //逆序对    private static int inversePairs(int[] arr,int length){        if (arr == null || arr.length<1) {            return 0;        }                    int[] copy = new int[length];    //辅助数组        for (int i=0;i<length;i++){            copy[i]=arr[i];        }        int count = inversePairsCore(arr,copy,0,length-1);   //计算逆序对        return count;    }    private static int inversePairsCore(int[] arr, int[] copy, int start, int end) {        //递归结束条件        if (start == end) {            //copy[start] = arr[start];  //分解成单元素后的处理            return 0;        }        int length = (end - start)/2;        int left = inversePairsCore(copy, arr, start, start+length);  //会一直执行到完成 (排序,注意参数1和2是反的)        int right = inversePairsCore(copy, arr,start+length+1, end);        /*合并,排序,并统计逆序对*/        //i初始化为前半段最后一个数字的坐标        int i = start+length;        //j初始化为后半段最后一个数字的坐标        int j = end;        int indexCopy = end;  //辅助数组的指针位置        int count = 0;        //可把i,j看出是两个指针        while (i >= start && j>=start+length+1){            if (arr[i]>arr[j]) {                copy[indexCopy--]=arr[i--];                count += j-start-length;            }else {                copy[indexCopy--]=arr[j--];            }        }        //右半边用尽,取左半边元素        for (;i>=start;i--){            copy[indexCopy--]=arr[i];        }        //左半边用尽,取右半边元素        for (;j>=start+length+1;j--){            copy[indexCopy--]=arr[j];        }        return left+right+count;    }    /*==================测试用例==============*/    private static void test(int[] data,int length,int expected){        if(inversePairs(data, length)== expected)            System.out.println("Passed.");        else            System.out.println("Failed.");    }    private static void test1() {        int[] data = {7,5,6,4};        int expected = 5;        test(data, data.length, expected);    }    private static void test2() {        int[] data = {6, 5, 4, 3, 2, 1};        int expected = 15;        test(data, data.length, expected);    }    private static void test3() {        int[] data = {1, 2, 3, 4, 5, 6};        int expected = 0;        test(data, data.length, expected);    }    private static void test4() {        int[] data = {1};        int expected = 0;           test(data, data.length, expected);    }    private static void test5() {        int[] data = {1, 2};        int expected = 0;           test(data, data.length, expected);    }    private static void test6() {        int[] data = {2, 1};        int expected = 1;           test(data, data.length, expected);    }    private static void test7() {        int[] data = {1, 2, 1, 2, 1};        int expected = 3;        test(data, data.length, expected);    }}

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有不妥当之处,麻烦告知:D