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Sort1

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要开始准备刷题啦！排序算法知识点总结，对数器的实现，master公式，小和问题，逆序对问题。Sort1包括冒泡排序，选择排序，插入排序，归并排序。

冒泡排序

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package basic_class_01;

import java.util.Arrays;

public class Code_00_BubbleSort {
    public static void bubbleSort(int[] arr) {
        if (arr == null || arr.length < 2) {
            return;
        }
        for (int e = arr.length - 1; e > 0; e--) {
            for (int i = 0; i < e; i++) {
                if (arr[i] > arr[i + 1]) {
                    swap(arr, i, i + 1);
                }
            }
        }
    }

    public static void swap(int[] arr, int i, int j) {
        //if (i==j){
        //	return;
    	//}
        //这里不写是因为i，j不可能相等
        arr[i] = arr[i] ^ arr[j];
        arr[j] = arr[i] ^ arr[j];
        arr[i] = arr[i] ^ arr[j];
    }

    //for test 对数器
    public static void comparator(int[] arr) {
        Arrays.sort(arr);
    }

    //for test
    public static int[] generateRandomArray(int maxSize, int maxValue) {
        int[] arr = new int[(int) ((maxSize + 1) * Math.random())];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = (int) ((maxValue + 1) * Math.random()) - (int) (maxValue * Math.random());
        }
        return arr;
    }

    //for test
    public static int[] copyArray(int[] arr) {
        if (arr == null) {
            return null;
        }
        int[] res = new int[arr.length];
        for (int i = 0; i < arr.length; i++) {
            res[i] = arr[i];
        }
        return res;
    }

    //for test
    public static boolean isEqual(int[] arr1, int[] arr2) {
        if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) {
            return false;
        }
        if (arr1 == null && arr2 == null) {
            return true;
        }
        if (arr1.length != arr2.length) {
            return false;
        }
        for (int i = 0; i < arr1.length; i++) {
            if (arr1[i] != arr2[i]) {
                return false;
            }
        }
        return true;
    }

    //for test
    public static void printArray(int[] arr) {
        if (arr == null) {
            return;
        }
        for (int i = 0; i < arr.length; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }

    //for test
    public static void main(String[] args) {
        int testTime = 500000;
        int maxSize = 100;
        int maxValue = 100;
        boolean succeed = true;
        for (int i = 0; i < testTime; i++) {
            int[] arr1 = generateRandomArray(maxSize, maxValue);
            int[] arr2 = copyArray(arr1);
            bubbleSort(arr1);
            comparator(arr2);
            if (!isEqual(arr1, arr2)) {
                succeed = false;
                break;
            }
        }
        System.out.println(succeed ? "Nice!" : "Fuck!");
        int[] arr = generateRandomArray(maxSize, maxValue);
        printArray(arr);
        bubbleSort(arr);
        printArray(arr);
    }
}
  • 时间复杂度 O(N^2)
  • 额外空间复杂度 O(1)

选择排序

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import java.util.Arrays;
public class Code_01_SelectionSort {
    public static void selectionSort(int[] arr) {
        if (arr == null || arr.length < 2) {
            return;
        }
        for (int i = 0; i < arr.length; i++) {
            int minIndex = i;
            for (int j = i + 1; j < arr.length; j++) {
                minIndex = arr[j] < arr[minIndex] ? j : minIndex;
            }
            swap(arr, i, minIndex);
        }
    }

    public static void swap(int[] arr, int i, int j) {
        int tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }
}
  • 时间复杂度 O(N^2)
  • 额外空间复杂度 O(1)

插入排序

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public class Code_02_InsertionSort {
    public static void insertionSort(int[] arr) {
        if (arr == null || arr.length < 2) {
            return;
        }
        for (int i = 1; i < arr.length; i++) {
            for (int j = i - 1; j >=0 && arr[j] > arr[j + 1]; j--) {
                swap(arr, j, j + 1);
            }
        }
    }

    public static void swap(int[] arr, int i, int j) {
        arr[i] = arr[i] ^ arr[j];
        arr[j] = arr[i] ^ arr[j];
        arr[i] = arr[i] ^ arr[j];
    }
}

差不多已经排序的数组，小数组可以用插入排序。完全有序O(N),逆序O(N^2)。

  • 时间复杂度 O(N^2) :
    • 顺序：只需比较N-1次，O(N)
    • 逆序：需比较 1+2+3+…+N-1次，O(N^2)
  • 额外空间复杂度 O(1)

归并排序

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//分治思想 递归实现
public class Code_03_MergeSort {
    public static void mergeSort(int[] arr) {
        if (arr == null || arr.length < 2) {
            return;
        }
        mergeSort(arr, 0, arr.length - 1);
    }

    public static void mergeSort(int[] arr, int l, int r) {
        if (l == r) {
            return;
        }
        int mid = l + ((r - l) >> 1);
        mergeSort(arr, l, mid);
        mergeSort(arr, mid + 1, r);
        merge(arr, l, mid, r);
    }

    public static void 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;
        while (p1 <= m && p2 <= r) {
            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];
        }
    }
}

master 公式

image-20200908172025543

  • 根据master公式，归并排序的时间复杂度：

    a=2,b=2, logb(a) = 1,d=1 —-> O(NlogN)

  • 额外空间复杂度 O(N)

小和问题和逆序对问题

image-20200909191623653

小和问题代码：

可以利用类似于归并排序的分治思想

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public class Code_12_SmallSum {
    public static int smallSum(int[] arr) {
        if (arr == null || arr.length < 2) {
            return 0;
        }
        return smallSum(arr, 0, arr.length - 1);
    }

    public static int smallSum(int[] arr, int l, int r) {
        if (l == r) {
            return 0;
        }
        int mid = l + ((r - l) >> 1);
        return smallSum(arr, l, mid) + smallSum(arr, mid + 1, r) + merge(arr, l, mid, r);

    }

    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] ? (arr[p1] * (r - p2 + 1)) : 0;
            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;
    }
}

逆序对问题也一样。