I 题目：
Given a sorted array, remove the duplicates in place such that each element appear onlyonce 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 ofnums being1 and2 respectively. It doesn’t matter what you leave beyond the new length.
解答：
设计两个指针：cur 和 pre。pre 正常遍历，，cur 则是记录更新后数组的边界。
注意：我的做法是比较当前元素与下一个元素，因此最后一个元素需要单独比较、处理。
class Solution {public:int removeDuplicates(vector<int>& nums) {int size = nums.size();if(size == 0) return 0;if(size == 1) return 1;int cur = 0;int pre = 0;for(; pre< size – 1; pre++){if(nums[pre] != nums[pre + 1]){nums[cur] = nums[pre];cur ++;}elsenums[cur] = nums[pre];}if (nums[pre] != nums[cur])nums[cur] = nums[pre];return cur + 1;}};
II 题目：
Follow up for "Remove Duplicates":What if duplicates are allowed at most twice?
For example,Given sorted array nums = [1,1,1,2,2,3],
Your function should return length = 5, with the first five elements ofnums being1,1, 2, 2 and3. It doesn’t matter what you leave beyond the new length.
解答：
和上一题同一个道理。只是这次多增加 count，记录当前是第几次重复上一个元素。另外，注意最后一个元素。
class Solution {public:int removeDuplicates(vector<int>& nums) {int count = 0;int size = nums.size();if (size <= 2) return size;int cur = 0;int pre = 0;for (; pre<size – 1; pre++) {if(nums[pre] != nums[pre + 1]) {if(count <= 1) {nums[cur] = nums[pre];cur++;}count = 0;}else if (nums[pre] == nums[pre + 1] && count <= 1) {nums[cur] = nums[pre];cur++;count++;}}if(count <= 1) {nums[cur] = nums[size – 1];cur++;}return cur;}};
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在前进的路上，主动搬开别人脚下的绊脚石，有时往往也是为自己铺路。
