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algorithm003
algorithm
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algorithm
/
Week_01
/
id_31
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LeetCode_26_31.cpp
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master
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algorithm
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Week_01
/
id_31
/
LeetCode_26_31.cpp
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#includ
e
<iostrea
m
>
#
include
<
vector
>
using
namespace
std
;
/*
给定一个排序数组,你需要在原地删除重复出现的元素,使得每个元素只出现一次,返回移除后数组的新长度。
不要使用额外的数组空间,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。
示例 1:
给定数组 nums = [1,1,2],
函数应该返回新的长度 2, 并且原数组 nums 的前两个元素被修改为 1, 2。
你不需要考虑数组中超出新长度后面的元素。
示例 2:
给定 nums = [0,0,1,1,1,2,2,3,3,4],
函数应该返回新的长度 5, 并且原数组 nums 的前五个元素被修改为 0, 1, 2, 3, 4。
你不需要考虑数组中超出新长度后面的元素。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
*/
namespace
version_1
{
class
Solution
{
public:
int
removeDuplicates
(vector<
int
>& nums) {
if
(nums.
empty
())
{
return
0
;
}
if
(nums.
size
() ==
1
)
{
return
1
;
}
int
result =
0
;
int
len = nums.
size
();
int
next =
1
;
int
cur =
0
;
for
(; next < len; next++)
{
if
(nums[cur] != nums[next])
{
cur++;
nums[cur] = nums[next];
}
}
return
cur +
1
;
}
};
}
using
namespace
version_1
;
int
main
(
int
argc,
char
*array[])
{
Solution mytest;
vector<
int
> test1{
1
,
1
,
2
};
int
result1 = mytest.
removeDuplicates
(test1);
vector<
int
> test2{
0
,
0
,
1
,
1
,
1
,
2
,
2
,
3
,
3
,
4
};
int
result2 = mytest.
removeDuplicates
(test2);
return
0
;
}
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