Question

我有一个

结构数组

typedef struct BinaryTreeNode {
    BSTElementType userID;
    BSTElementType count;
    struct BinaryTreeNode *left;
    struct BinaryTreeNode *right;
}BinaryTreeNode;

typedef BinaryTreeNode BSTNode;

我希望按升序合并排序数组。但是，当我执行合并时，没有任何变化。这是我用来创建struct数组的代码，以及MergeSort的函数调用。 maxUsers是我从转换二叉树中节点数得到的int，它应该是数组的最大数量。

BSTNode *arr = (BSTNode *)malloc(maxUsers * sizeof(BSTNode));

MergeSort(arr, 0, maxUsers);

void Merge(BSTNode *arr, int low, int mid, int high) {
    int mergedSize = high - low + 1;
    BSTNode *temp = (BSTNode *)malloc(mergedSize * sizeof(BSTNode));
    int mergePos = 0;
    int leftPos = 0;
    int rightPos = 0;

    leftPos = low;
    rightPos = mid + 1;

    //printf("Starting Merge\n");

    while (leftPos <= high && rightPos <= mid) {
        if (arr[leftPos].count < arr[rightPos].count) {
            temp[mergePos].userID = arr[leftPos].userID;
            temp[mergePos].count = arr[leftPos].count;
            ++leftPos;
        }
        else {
            temp[mergePos].userID = arr[rightPos].userID;
            temp[mergePos].count = arr[rightPos].count;
            ++rightPos;
        }
        ++mergePos;
    }

    while (leftPos <= mid) {
        temp[mergePos].userID = arr[leftPos].userID;
        temp[mergePos].count = arr[leftPos].count;
        ++leftPos;
        ++mergePos;
    }

    while (rightPos <= high) {
        temp[mergePos].userID = arr[rightPos].userID;
        temp[mergePos].count = arr[rightPos].count;
        ++rightPos;
        ++mergePos;
    }

    for (mergePos = 0; mergePos < mergedSize; ++mergePos) {
        arr[low + mergePos].userID = temp[mergePos].userID;
        arr[low + mergePos].count = temp[mergePos].count;
    }
}

void MergeSort(BSTNode *arr, int low, int high) {
    int mid = 0;

    if (low < high) {
        mid = (low + high) / 2;

        MergeSort(arr, low, mid);
        MergeSort(arr, mid + 1, high);

        Merge(arr, low, mid, high);
    }
}

非常感谢任何提示或提示！

编辑：当我尝试编写一些printf语句时，我注意到这些值是负面的。但是存储在结构中的值是正数。出现此错误的原因是什么？

printf("Left Pos: %d, Right Pos: %d\n", leftPos, rightPos);
    while (leftPos <= mid && rightPos <= high) {
        printf("Left Pos count: %d, Right Pos count: %d\n", arr[leftPos].count, arr[rightPos].count);
        if (arr[leftPos].count < arr[rightPos].count) {
            temp[mergePos].userID = arr[leftPos].userID;
            temp[mergePos].count = arr[leftPos].count;
            ++leftPos;
        }

Answer 1

需要做出两个重要的变化：

在Merge中，您有while (leftPos <= high && rightPos <= mid)，但需要while (leftPos <= mid && rightPos <= high)（反转左/右比较值）。
在main() - 或者至少是调用MergeSort()的代码中，您有MergeSort(arr, 0, maxUsers);，但您需要MergeSort(arr, 0, maxUsers-1);因为传递的值必须是arr[lo] arr[hi]和maxUsers都是数组中的有效条目，当您通过arr[maxUsers]时，您告诉它free(temp)在有效时不会有效。

第三个重要更改是在退出MergeSort()之前添加rand()。

我还选择用简单的单行结构分配来替换多行分配。

这是代码的调试版本，但调试打印已全部注释掉。当我解决问题时，它全部活跃起来。在我调试时，即使我使用srand()生成数据，我也没有故意使用srand(time(0));，因此我在每次运行时都使用相同的数据。当我确信代码是干净的时候，我使用#include <assert.h> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <time.h> typedef int BSTElementType; typedef struct BinaryTreeNode { BSTElementType userID; BSTElementType count; struct BinaryTreeNode *left; struct BinaryTreeNode *right; } BinaryTreeNode; typedef BinaryTreeNode BSTNode; static void PrintArray(const char *tag, int n, BSTNode *array) { printf("%s:\n", tag); for (int i = 0; i < n; i++) printf("%2d: u = %4d, c = %2d\n", i, array[i].userID, array[i].count); putchar('\n'); fflush(stdout); } static void Merge(BSTNode *arr, int low, int mid, int high) { int mergedSize = high - low + 1; BSTNode *temp = (BSTNode *)malloc(mergedSize * sizeof(BSTNode)); int mergePos = 0; int leftPos = low; int rightPos = mid + 1; //printf("-->> %s: lo = %d, md = %d, hi = %d, sz = %d\n", __func__, low, mid, high, mergedSize); //printf("L = %d, M = %d; R = %d, H = %d\n", leftPos, mid, rightPos, high); while (leftPos <= mid && rightPos <= high) // Key change { //printf("a[%d].c = %d <=> a[%d].c = %d\n", leftPos, arr[leftPos].count, // rightPos, arr[rightPos].count); if (arr[leftPos].count < arr[rightPos].count) { //printf("L1: a[%d].c = %d\n", leftPos, arr[leftPos].count); temp[mergePos++] = arr[leftPos++]; } else { //printf("R1: a[%d].c = %d\n", rightPos, arr[rightPos].count); temp[mergePos++] = arr[rightPos++]; } //printf("L = %d, M = %d; R = %d, H = %d\n", leftPos, mid, rightPos, high); } while (leftPos <= mid) { //printf("L2: a[%d].c = %d\n", leftPos, arr[leftPos].count); temp[mergePos++] = arr[leftPos++]; } while (rightPos <= high) { //printf("R2: a[%d].c = %d\n", rightPos, arr[rightPos].count); temp[mergePos++] = arr[rightPos++]; } assert(mergePos == mergedSize); //PrintArray("merged", mergedSize, temp); for (mergePos = 0; mergePos < mergedSize; ++mergePos) arr[low + mergePos] = temp[mergePos]; free(temp); // Key change //printf("<<-- %s: lo = %d, md = %d, hi = %d, sz = %d\n", __func__, low, mid, high, mergedSize); } static void MergeSort(BSTNode *arr, int low, int high) { if (low < high) { int mid = (low + high) / 2; //printf("-->> %s: lo = %d, md = %d, hi = %d\n", __func__, low, mid, high); MergeSort(arr, low, mid); MergeSort(arr, mid + 1, high); Merge(arr, low, mid, high); //printf("<<-- %s: lo = %d, md = %d, hi = %d\n", __func__, low, mid, high); } } int main(void) { /* Unstable when testing */ //srand(time(0)); //int maxUsers = rand() % 20 + 5; /* Stable while debugging */ int maxUsers = 10; BSTNode *arr = (BSTNode *)malloc(maxUsers * sizeof(BSTNode)); for (int i = 0; i < maxUsers; i++) { arr[i].userID = rand() % 100 * 100; arr[i].count = rand() % 10; arr[i].left = 0; arr[i].right = 0; } PrintArray("Before", maxUsers, arr); MergeSort(arr, 0, maxUsers - 1); // Key change PrintArray("After", maxUsers, arr); free(arr); return 0; }并改变了数组的大小，但是在调试时，稳定性很重要。

Merge()

关键的调试观察之一是Before: 0: u = 7900, c = 9 1: u = 3900, c = 5 2: u = 5500, c = 3 3: u = 4300, c = 4 4: u = 3600, c = 6 5: u = 900, c = 2 6: u = 5300, c = 9 7: u = 6300, c = 1 8: u = 7900, c = 8 9: u = 4400, c = 3 10: u = 9400, c = 0 After: 0: u = 9400, c = 0 1: u = 6300, c = 1 2: u = 900, c = 2 3: u = 4400, c = 3 4: u = 5500, c = 3 5: u = 4300, c = 4 6: u = 3900, c = 5 7: u = 3600, c = 6 8: u = 7900, c = 8 9: u = 5300, c = 9 10: u = 7900, c = 9中的主循环从未进入过;这是对第一个问题的强烈暗示。 Oddball数据是第二个问题的强烈暗示。主要数据都在很好的范围内控制。

示例输出（YMMV，因为您可能没有使用完全相同的PRNG）：

Before:
 0: u = 3300, c =  1
 1: u =  200, c =  8
 2: u = 7500, c =  6
 3: u = 2700, c =  8
 4: u = 8300, c =  3
 5: u = 6900, c =  3
 6: u =  400, c =  3
 7: u = 1100, c =  6
 8: u = 3600, c =  5
 9: u = 2100, c =  7
10: u = 9400, c =  9
11: u = 7300, c =  0
12: u = 1800, c =  4
13: u = 8200, c =  9
14: u = 8400, c =  4
15: u = 9600, c =  0
16: u = 4400, c =  7
17: u = 2900, c =  0
18: u = 4300, c =  4
19: u = 6500, c =  9
20: u = 6800, c =  6
21: u = 8000, c =  2
22: u = 6000, c =  5

After:
 0: u = 2900, c =  0
 1: u = 9600, c =  0
 2: u = 7300, c =  0
 3: u = 3300, c =  1
 4: u = 8000, c =  2
 5: u =  400, c =  3
 6: u = 6900, c =  3
 7: u = 8300, c =  3
 8: u = 4300, c =  4
 9: u = 8400, c =  4
10: u = 1800, c =  4
11: u = 6000, c =  5
12: u = 3600, c =  5
13: u = 6800, c =  6
14: u = 1100, c =  6
15: u = 7500, c =  6
16: u = 4400, c =  7
17: u = 2100, c =  7
18: u = 2700, c =  8
19: u =  200, c =  8
20: u = 6500, c =  9
21: u = 8200, c =  9
22: u = 9400, c =  9

样本运行（随机数据量）：

Before:
 0: u =  700, c =  9
 1: u = 7300, c =  8
 2: u = 3000, c =  2
 3: u = 4400, c =  8
 4: u = 2300, c =  9
 5: u = 4000, c =  5
 6: u = 9200, c =  2
 7: u = 8700, c =  3
 8: u = 2700, c =  9
 9: u = 4000, c =  2

-->> MergeSort: lo = 0, md = 4, hi = 9
-->> MergeSort: lo = 0, md = 2, hi = 4
-->> MergeSort: lo = 0, md = 1, hi = 2
-->> MergeSort: lo = 0, md = 0, hi = 1
-->> Merge: lo = 0, md = 0, hi = 1, sz = 2
L = 0, M = 0; R = 1, H = 1
a[0].c = 9 <=> a[1].c = 8
R1: a[1].c = 8
L = 0, M = 0; R = 2, H = 1
L2: a[0].c = 9
<<-- Merge: lo = 0, md = 0, hi = 1, sz = 2
<<-- MergeSort: lo = 0, md = 0, hi = 1
-->> Merge: lo = 0, md = 1, hi = 2, sz = 3
L = 0, M = 1; R = 2, H = 2
a[0].c = 8 <=> a[2].c = 2
R1: a[2].c = 2
L = 0, M = 1; R = 3, H = 2
L2: a[0].c = 8
L2: a[1].c = 9
<<-- Merge: lo = 0, md = 1, hi = 2, sz = 3
<<-- MergeSort: lo = 0, md = 1, hi = 2
-->> MergeSort: lo = 3, md = 3, hi = 4
-->> Merge: lo = 3, md = 3, hi = 4, sz = 2
L = 3, M = 3; R = 4, H = 4
a[3].c = 8 <=> a[4].c = 9
L1: a[3].c = 8
L = 4, M = 3; R = 4, H = 4
R2: a[4].c = 9
<<-- Merge: lo = 3, md = 3, hi = 4, sz = 2
<<-- MergeSort: lo = 3, md = 3, hi = 4
-->> Merge: lo = 0, md = 2, hi = 4, sz = 5
L = 0, M = 2; R = 3, H = 4
a[0].c = 2 <=> a[3].c = 8
L1: a[0].c = 2
L = 1, M = 2; R = 3, H = 4
a[1].c = 8 <=> a[3].c = 8
R1: a[3].c = 8
L = 1, M = 2; R = 4, H = 4
a[1].c = 8 <=> a[4].c = 9
L1: a[1].c = 8
L = 2, M = 2; R = 4, H = 4
a[2].c = 9 <=> a[4].c = 9
R1: a[4].c = 9
L = 2, M = 2; R = 5, H = 4
L2: a[2].c = 9
<<-- Merge: lo = 0, md = 2, hi = 4, sz = 5
<<-- MergeSort: lo = 0, md = 2, hi = 4
-->> MergeSort: lo = 5, md = 7, hi = 9
-->> MergeSort: lo = 5, md = 6, hi = 7
-->> MergeSort: lo = 5, md = 5, hi = 6
-->> Merge: lo = 5, md = 5, hi = 6, sz = 2
L = 5, M = 5; R = 6, H = 6
a[5].c = 5 <=> a[6].c = 2
R1: a[6].c = 2
L = 5, M = 5; R = 7, H = 6
L2: a[5].c = 5
<<-- Merge: lo = 5, md = 5, hi = 6, sz = 2
<<-- MergeSort: lo = 5, md = 5, hi = 6
-->> Merge: lo = 5, md = 6, hi = 7, sz = 3
L = 5, M = 6; R = 7, H = 7
a[5].c = 2 <=> a[7].c = 3
L1: a[5].c = 2
L = 6, M = 6; R = 7, H = 7
a[6].c = 5 <=> a[7].c = 3
R1: a[7].c = 3
L = 6, M = 6; R = 8, H = 7
L2: a[6].c = 5
<<-- Merge: lo = 5, md = 6, hi = 7, sz = 3
<<-- MergeSort: lo = 5, md = 6, hi = 7
-->> MergeSort: lo = 8, md = 8, hi = 9
-->> Merge: lo = 8, md = 8, hi = 9, sz = 2
L = 8, M = 8; R = 9, H = 9
a[8].c = 9 <=> a[9].c = 2
R1: a[9].c = 2
L = 8, M = 8; R = 10, H = 9
L2: a[8].c = 9
<<-- Merge: lo = 8, md = 8, hi = 9, sz = 2
<<-- MergeSort: lo = 8, md = 8, hi = 9
-->> Merge: lo = 5, md = 7, hi = 9, sz = 5
L = 5, M = 7; R = 8, H = 9
a[5].c = 2 <=> a[8].c = 2
R1: a[8].c = 2
L = 5, M = 7; R = 9, H = 9
a[5].c = 2 <=> a[9].c = 9
L1: a[5].c = 2
L = 6, M = 7; R = 9, H = 9
a[6].c = 3 <=> a[9].c = 9
L1: a[6].c = 3
L = 7, M = 7; R = 9, H = 9
a[7].c = 5 <=> a[9].c = 9
L1: a[7].c = 5
L = 8, M = 7; R = 9, H = 9
R2: a[9].c = 9
<<-- Merge: lo = 5, md = 7, hi = 9, sz = 5
<<-- MergeSort: lo = 5, md = 7, hi = 9
-->> Merge: lo = 0, md = 4, hi = 9, sz = 10
L = 0, M = 4; R = 5, H = 9
a[0].c = 2 <=> a[5].c = 2
R1: a[5].c = 2
L = 0, M = 4; R = 6, H = 9
a[0].c = 2 <=> a[6].c = 2
R1: a[6].c = 2
L = 0, M = 4; R = 7, H = 9
a[0].c = 2 <=> a[7].c = 3
L1: a[0].c = 2
L = 1, M = 4; R = 7, H = 9
a[1].c = 8 <=> a[7].c = 3
R1: a[7].c = 3
L = 1, M = 4; R = 8, H = 9
a[1].c = 8 <=> a[8].c = 5
R1: a[8].c = 5
L = 1, M = 4; R = 9, H = 9
a[1].c = 8 <=> a[9].c = 9
L1: a[1].c = 8
L = 2, M = 4; R = 9, H = 9
a[2].c = 8 <=> a[9].c = 9
L1: a[2].c = 8
L = 3, M = 4; R = 9, H = 9
a[3].c = 9 <=> a[9].c = 9
R1: a[9].c = 9
L = 3, M = 4; R = 10, H = 9
L2: a[3].c = 9
L2: a[4].c = 9
<<-- Merge: lo = 0, md = 4, hi = 9, sz = 10
<<-- MergeSort: lo = 0, md = 4, hi = 9
After:
 0: u = 4000, c =  2
 1: u = 9200, c =  2
 2: u = 3000, c =  2
 3: u = 8700, c =  3
 4: u = 4000, c =  5
 5: u = 4400, c =  8
 6: u = 7300, c =  8
 7: u = 2700, c =  9
 8: u = 2300, c =  9
 9: u =  700, c =  9

启用调试，固定大小的运行：

import pandas as pd
noTEU = pd.DataFrame() # empty database
index_TEU = 0
for vessel in list:
    if condition is fullfilled:
        imo_vessel = pd.DataFrame({'imo': vessel}, index=[index_TEU])   
        noTEU.append(imo_vessel) # I want here to add an element to my database
        index_TEU = index_TEU + 1

代码在Mac OS X 10.11.6上的GCC 6.1.0和Valgrind 3.12-SVN下编译和运行。

如何合并排序结构数组

1 个答案: