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]
.
删除数组中的重复元素。
思路,设置两个下标i,j,使用j遍历数组,如果A[j]和A[j+1]不相等,则把A[i]赋给A[j],否则跳过位置j,不进行赋值操作,相当于删除了A[j],遍历完成后还需要将A[n-1]赋给A[i],因为A[n-1]要么本来就不重复,要么重复但是其前面的重复数字都被删除了,不管哪种情况都需要保存A[n-1]。
算法时间复杂度O(n),空间复杂度O(1)。
accepted solution:
class Solution {
public:
int removeDuplicates(int A[], int n)
{
if(n==0)
return 0;
int i=0;
for(int j=0;j<n-1;j++)
{
if(A[j]!=A[j+1])
A[i++]=A[j];
}
A[i++]=A[n-1];
return i;
}
};