关于#数组#的问题，如何解决？

Question

：给你一个 升序排列 的数组 nums ，请你 原地 删除重复出现的元素，使每个元素 只出现一次 ，返回删除后数组的新长度。元素的 相对顺序 应该保持 一致 。

由于在某些语言中不能改变数组的长度，所以必须将结果放在数组nums的第一部分。更规范地说，如果在删除重复项之后有 k 个元素，那么 nums 的前 k 个元素应该保存最终结果。
将最终结果插入 nums 的前 k 个位置后返回 k 。
不要使用额外的空间，你必须在 原地 修改输入数组 并在使用 O(1) 额外空间的条件下完成。

作者：力扣 (LeetCode)
链接：https://leetcode.cn/leetbook/read/top-interview-questions-easy/x2gy9m/

```c
int removeDuplicates(int* nums, int numsSize){
 if (numsSize <= 1) {
        return numsSize;
    }

    int left = 0,right = 1;
    while (left+1 < numsSize) {
        if (nums[left] != nums[right]) {
           nums[++left] = nums[right++];
        }else{ right++;}
     
    }
     return  ++left;
}

###### 类似于双指针的思想：两个指针，右指针一直往右。1.当左指针值不等于右指针值，左指针右移一位，并把右指针的值赋给此时的左指针，然后右指针右移一位；2.当左指针的值等于右指针的值，左指针不动，右指针右移。最后返回left+1的值。


###### 执行出错信息：
=================================================================
==42==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x60200000001c at pc 0x55f0875d8db3 bp 0x7ffea3e09490 sp 0x7ffea3e09480
READ of size 4 at 0x60200000001c thread T0
    #2 0x7f39894ec082 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x24082)
0x60200000001c is located 0 bytes to the right of 12-byte region [0x602000000010,0x60200000001c)
allocated by thread T0 here:
    #0 0x7f398a134808 in __interceptor_malloc ../../../../src/libsanitizer/asan/asan_malloc_linux.cc:144
    #3 0x7f39894ec082 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[04]fa fa fa fa 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
==42==ABORTING

Answer 1

移位覆盖