算法之美：探究左右元素和的差值

news2025/2/27 0:50:40

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本篇博客会讲解力扣“2574. 左右元素和的差值”的解题思路，这是题目链接。

先来审题：
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以下是输出示例：

以下是提示：

本题的关键在于，“左和”和“右和”是如何变化的。下面我通过代码来演示。

一上来，我们应该求下表为0的元素的“左和”和“右和”。

int* leftRigthDifference(int* nums, int numsSize, int* returnSize){
    // 下标0元素的左和
    int left = 0;
    // 下标0元素的右和
    int right = 0;
    for (int i = 1; i < numsSize; ++i)
    {
        right += nums[i];
    }
}

接下来，开辟结果数组。

int* leftRigthDifference(int* nums, int numsSize, int* returnSize){
    // 下标0元素的左和
    int left = 0;
    // 下标0元素的右和
    int right = 0;
    for (int i = 1; i < numsSize; ++i)
    {
        right += nums[i];
    }

    int* ret = (int*)malloc(sizeof(int) * numsSize);
    *returnSize = numsSize;
}

遍历结果数组，把结果放进去。

int* leftRigthDifference(int* nums, int numsSize, int* returnSize){
    // 下标0元素的左和
    int left = 0;
    // 下标0元素的右和
    int right = 0;
    for (int i = 1; i < numsSize; ++i)
    {
        right += nums[i];
    }

    int* ret = (int*)malloc(sizeof(int) * numsSize);
    *returnSize = numsSize;

    for (int i = 0; i < numsSize; ++i)
    {
        ret[i] = abs(left - right);
    }

    return ret;
}

下面是关键：“左和”和“右和”是如何变化的？

对于“左和”，下标为1的元素的“左和”比起下标为0的元素的“左和”，多加了下标为0的元素，所以“左和”的变化规律是：left += nums[i];。
对于“右和”，下标为1的元素的“右和”比起下标为0的元素的“右和”，需要减去下标为1的元素，所以“右和”的变化规律是：right -= nums[i + 1];。

写进代码：

int* leftRigthDifference(int* nums, int numsSize, int* returnSize){
    // 下标0元素的左和
    int left = 0;
    // 下标0元素的右和
    int right = 0;
    for (int i = 1; i < numsSize; ++i)
    {
        right += nums[i];
    }

    int* ret = (int*)malloc(sizeof(int) * numsSize);
    *returnSize = numsSize;

    for (int i = 0; i < numsSize; ++i)
    {
        ret[i] = abs(left - right);
        left += nums[i];
        right -= nums[i + 1];
    }

    return ret;
}

但是这样过不了，错误信息如下：

=================================================================
==20==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x602000000020 at pc 0x55f869e36270 bp 0x7fff04e380c0 sp 0x7fff04e380b0
READ of size 4 at 0x602000000020 thread T0
    #2 0x7f4221987082 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x24082)
0x602000000020 is located 0 bytes to the right of 16-byte region [0x602000000010,0x602000000020)
allocated by thread T0 here:
    #0 0x7f42225cf808 in __interceptor_malloc ../../../../src/libsanitizer/asan/asan_malloc_linux.cc:144
    #3 0x7f4221987082 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x24082)
Shadow bytes around the buggy address:
  0x0c047fff7fb0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fc0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fd0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fe0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7ff0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
=>0x0c047fff8000: fa fa 00 00[fa]fa 00 00 fa fa fa fa fa fa fa fa
  0x0c047fff8010: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8020: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8030: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8040: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8050: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
Shadow byte legend (one shadow byte represents 8 application bytes):
  Addressable:           00
  Partially addressable: 01 02 03 04 05 06 07 
  Heap left redzone:       fa
  Freed heap region:       fd
  Stack left redzone:      f1
  Stack mid redzone:       f2
  Stack right redzone:     f3
  Stack after return:      f5
  Stack use after scope:   f8
  Global redzone:          f9
  Global init order:       f6
  Poisoned by user:        f7
  Container overflow:      fc
  Array cookie:            ac
  Intra object redzone:    bb
  ASan internal:           fe
  Left alloca redzone:     ca
  Right alloca redzone:    cb
  Shadow gap:              cc
==20==ABORTING

这是为什么呢？其实，当计算到下标为size-1的元素的“右和”时，执行right -= nums[i+1];，i+1已经越界了。所以最后一个右和要单独算。

int* leftRigthDifference(int* nums, int numsSize, int* returnSize){
    // 下标0元素的左和
    int left = 0;
    // 下标0元素的右和
    int right = 0;
    for (int i = 1; i < numsSize; ++i)
    {
        right += nums[i];
    }

    int* ret = (int*)malloc(sizeof(int) * numsSize);
    *returnSize = numsSize;

    for (int i = 0; i < numsSize - 1; ++i)
    {
        ret[i] = abs(left - right);
        left += nums[i];
        right -= nums[i + 1];
    }

    // 防止越界，最后一个数据单独计算
    ret[numsSize - 1] = abs(left);

    return ret;
}