QUESTION:
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 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.
思路:
有序数组,顺序比较相邻数字,重复删除,注意循环中使用erase导致size变小,i--调整标号
CODE:class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.size()==0) return 0;
if(nums.size()==1) return 1;
for(int i = 0 ;i <nums.size() - 1; i++){
if(nums[i] == nums[i+1]){
nums.erase(nums.begin()+i);
i--;
}
}
int mylength = nums.size();
return mylength;
}
};
Your runtime beats 11.50% of cpp submissions.