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 by modifying the input array in-place with O(1) extra memory.
Example:
Given nums = [1,1,2], Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn't matter what you leave beyond the new length.
class Solution {
public:
int removeDuplicates(vector<int>& a) {
if(a.size()==) return ;
int i = ;
int j = ;
for (;j < a.size();++j) {
if(a[i]!=a[j]) {
i++;
}
a[i] = a[j];
}
return i+;
}
};
与删除排序链表中的重复元素类似，利用排序的特性，如果后一个元素大于当前元素，则不是重复的数字
class Solution:
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums)< 2:
return len(nums)
j = 0
for i in range(1,len(nums)):
if(nums[i]>nums[j]):
j+=1
nums[j]=nums[i] return j+1
20180507
class Solution {
public int removeDuplicates(int[] nums) {
int m = 0;
for(int i = 0;i<nums.length;i++){
if(m<1||nums[i]>nums[m-1])
nums[m++] = nums[i];
}
return m;
}
}