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.
Error Solution: runtime error
1 class Solution { 2 public: 3 int removeDuplicates(vector<int>& nums) { 4 if(nums.size()<2)return nums.size(); 5 for(vector<int>::iterator iter1=nums.begin(),iter2=nums.begin()+1;iter2!=nums.end();iter1++,iter2++){ 6 if(*iter1==*iter2)nums.erase(iter1);//这里erase之后iter1所指向的元素已经不存在了，故iter1已经无效 7 } 8 return nums.size(); 9 } 10 };
Solution: "It doesn't matter what you leave beyond the new length."主要是理解这句话，原vector不需要删除元素，用length保存当前下标，直到找到下一个不一样的元素再++
1 class Solution { 2 public: 3 int removeDuplicates(vector<int>& nums) { 4 if(nums.size()<2)return nums.size(); 5 int length=1; 6 for(int i=1;i<nums.size();i++){ 7 if(nums[i]!=nums[i-1]) 8 nums[length++]=nums[i]; 9 else 10 continue; 11 } 12 return length; 13 } 14 };