旋转数组LeetCode(189)

Rotate Array LeetCode (189)(旋转数组LeetCode(189))
本文介绍了旋转数组LeetCode(189)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

问题如下:

给定一个数组,将该数组向右旋转k步,其中k为非负数。

以下是我的代码:

 class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int r =nums.size()-k;
        vector<int>::iterator it;
        it = nums.begin();
        for(int i=0;i<r;i++){
         nums.push_back(nums[0]);
          nums.erase(it);
      }
    }
};

测试用例1:
输入:数字=[1,2,3,4,5,6,7],k=3
输出:[5,6,7,1,2,3,4]

在这里,我的代码已成功编译,并且提供了正确的解决方案。

测试用例2:
输入:数字=[-1,-100,3,99],k=2
输出:[3,99,-1,-100]

在这里,所有问题都开始了,我的代码显示以下错误:

=================================================================
==32==ERROR: AddressSanitizer: heap-use-after-free on address 0x602000000094 at pc 0x0000003189cf bp 0x7ffe0e44adf0 sp 0x7ffe0e44a5b8
READ of size 68719476672 at 0x602000000094 thread T0
    #5 0x7f15fa2470b2  (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
0x6020000000a0 is located 0 bytes to the right of 16-byte region [0x602000000090,0x6020000000a0)
freed by thread T0 here:
    #4 0x7f15fa2470b2  (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
previously allocated by thread T0 here:
    #6 0x7f15fa2470b2  (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
Shadow bytes around the buggy address:
  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 fd fa fa fa fd fa fa fa fd fa fa fa fd fa
=>0x0c047fff8010: fa fa[fd]fd 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
  0x0c047fff8060: 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
==32==ABORTING

我就是此错误,请帮助。

推荐答案

代码崩溃,因为it调用push_back后会invalidated,可以直接调用begin修复。

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int r =nums.size()- (k % nums.size());
        for(int i=0;i<r;i++){
            nums.push_back(nums[0]);
            nums.erase(nums.begin());
        }
    }
};

代码中的算法与右旋转无关,您使用的是左旋转。对于右旋转,需要将最后一个元素插入字体,然后擦除最后一个元素,我们还需要对向量的长度进行模型化,以避免不连续的旋转,否则会有浪费,公认的代码可能如下所示:

class Solution {
 public:
  void rotate(vector<int>& nums, int k) {
    int r = k % nums.size();
    for (int i = 0; i < r; i++) {
      nums.insert(nums.begin(), nums.back());
      nums.erase(std::prev(nums.end()));
    }
  }
};

我们还可以调用STL算法:rotate,要向右旋转,我们需要在这里使用反向迭代器:

class Solution {
 public:
  void rotate(vector<int>& nums, int k) {
    int r = k % nums.size();
    std::rotate(nums.rbegin(), nums.rbegin() + r, nums.rend());
  }
};

您的算法会导致向量元素在每次插入到前面时都会移位,效率不高,想想我们有一个很大的向量,移位所有元素都会成为瓶颈。

STL版本可能有一些其他性能增强,因为它可以成批移动元素范围,而不是逐个交换元素。

我们也可以调用三次std::reverse来实现我们自己的rotate

class Solution {
 public:
  void rotate(vector<int>& nums, int k) {
    int r = k % nums.size();
    std::reverse(nums.begin(), nums.end());
    std::reverse(nums.begin(), nums.begin() + r);
    std::reverse(nums.begin() + r, nums.end());
  }
};

最后两个版本比前两个版本快很多,建议使用。

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