描述
Given a sorted array, remove the duplicates in place such that each element appear only once
and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example, Given input array
A = [1,1,2],
Your function should return length = 2, and A is now
[1,2].
分析
无代码
<span style="font-size:14px;">#include <iostream>
#include <list>
#include <algorithm>
#include <iterator>
#include <functional>
using namespace std;
class Solution
{
public:
// 使用 STL,时间复杂度 O(n),空间复杂度 O(1)
static int removeDuplicatesBySTL(int A[], int n)
{
return distance(A, unique(A, A + n));
}
//遍历解法
//时间复杂度 O(n),空间复杂度 O(1)
static int removeDuplicatesByScan(int A[], int n)
{
if (n == 0) return 0;
int index = 0;
for (int i = 1; i < n; i++) {
if (A[index] != A[i])
A[++index] = A[i];
}
return index + 1;
}
};
int main()
{
int A[] = {1,1,2};
cout << "Length:" << Solution::removeDuplicatesBySTL(A, 3) << endl;
cout << "Length:" << Solution::removeDuplicatesByScan(A, 3) << endl;
return 0;
}</span>