声明：本文资料来自于 牛客网（算法初级笔记1）

1 小和问题

public static void main(String[] args){
        int[] arr = Util.generateRandomArray(8,10);
        smallSum(arr);
    }

    public static void smallSum(int[] arr){
        if (arr == null || arr.length < 2) {
            return;
        }
        System.out.println("原数组");
        Util.printArray(arr);
        System.out.println("");
        int smallSum = deal(arr,0,arr.length-1);
        System.out.println("处理后的数组：");
        Util.printArray(arr);
        System.out.println("小和为："+smallSum);
    }

    public static int deal(int[] arr,int left,int right){
        if (left == right){
            return 0;
        }
        int middle = left + ((right-left)>>1);
        return  deal(arr,left,middle) + deal(arr,middle+1,right)+merge(arr,left,middle,right);
    }

    public static int merge(int[] arr,int left,int middle,int right){
        int[] help = new int[right-left+1];
        int sum = 0;
        int index = 0;
        int p1=left;
        int p2=middle+1;
        while (p1 <= middle && p2 <= right){
            if (arr[p1] < arr[p2]){
                System.out.println((right-p2+1)+"个"+arr[p1]+" \n");//打印加数
            }
            sum += arr[p1] < arr[p2]?  (right-p2+1)*arr[p1]:0;
            help[index++] = arr[p1] < arr[p2]? arr[p1++] : arr[p2++];
        }
        while (p1 <= middle){
            help[index++] = arr[p1++];
        }
        while (p2 <= right){
            help[index++] = arr[p2++];
        }
        for (int i = 0;i < help.length;i++){
            arr[left+i] = help[i];
        }
        return sum;
    }

随机测试结果：

2 逆序对问题

在一个数组中，左边的数如果比右边的数大，则折两个数构成一个逆序对，请打印所有逆序对。

public static void main(String[] args){
        int[] arr = Util.generateRandomArray(8,10);
        System.out.println("原始数组："+Arrays.toString(arr));
        nixudui(arr);
        System.out.print("处理后数组："+Arrays.toString(arr));
    }

    public static void nixudui(int[] arr){
        deal(arr,0,arr.length-1);
    }

    public static void deal(int[] arr,int left,int right){
        if (left == right){
            return;
        }
        int middle = left + ((right-left)>>1);
        deal(arr,left,middle);
        deal(arr,middle+1,right);
        merge(arr,left,middle,right);

    }
    public static void merge(int[] arr, int left,int middle, int right){
        int p1=left;
        int p2 = middle+1;
        int i = 0;
        int[] help = new int[right-left+1];
        while (p1 <= middle && p2 <= right){
            if (arr[p1] > arr[p2]){
                for(int j = 0;j < middle-p1+1;j++){
                System.out.println("逆序对："+arr[p1+j]+"  "+arr[p2]);
                }
                help[i++] = arr[p2++];
            }else {
                help[i++] = arr[p1++];
            }
        }
        while (p1 <= middle){
            help[i++] = arr[p1++];
        }
        while (p2 <= right){
            help[i++] = arr[p2++];
        }
        for ( i = 0;i < help.length;i++){
            arr[left+i] = help[i];
        }
    }

2个的时间复杂度为O(N * log N),空间复杂度为O(N)