26. Remove Duplicates from Sorted Array
Given a sorted array nums, 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 1:
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 returned length.
双指针，i指向没有duplicates的字段的末尾（不包含最后一个数），j向后遍历寻找不重复的数，找到就加到i的位置上。两种代码实现：
第一种是我自己的写的：
int removeDuplicates(vector<int>& nums) { if (nums.size() == 0) { return 0; } int i = 0, j = 0, rightBound = (int)nums.size() - 1; nums.push_back(nums[0] - 1); while (j <= rightBound) { while (j <= rightBound && nums[j] != nums[j+1]) { nums[i++] = nums[j++]; } while (j <= rightBound && nums[j] == nums[j+1]) { j++; } } return i; }第二种是discuss里面别人的做法，简洁很多：
int removeDuplicates(vector<int>& nums) { int i = 0; for (int n : nums) if (!i || n > nums[i-1]) nums[i++] = n; return i; }两种里面i都指向可以直接插入的位置。
80. Remove Duplicates from Sorted Array II
Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice 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 1:
Given nums = [1,1,1,2,2,3],Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.It doesn't matter what you leave beyond the returned length.
通用是双指针的思路，同样有两种实现，
第一种：
// Idea: two pointers // j runs faster then i. i tracks the right-side boudary of result array, j runs ahead to move valid nums to nums[i] // [0, 0, 1, 3, 3, 7, 7, 7, 7, 8, 9, 10, ...] // i j int removeDuplicates(vector<int>& nums) { if (nums.size() == 0) { return 0; } int i = 0, j = 0, rightBound = (int)nums.size() - 1; nums.push_back(nums[0] - 1); while (j <= rightBound) { while (j <= rightBound && nums[j] != nums[j+1]) { nums[i++] = nums[j++]; } if (j < rightBound) { nums[i++] = nums[j++]; } while (j <= rightBound && nums[j] == nums[j+1]) { j++; } } return i; }第二种：
int removeDuplicates(vector<int>& nums) { int i = 0; for (int n : nums) if (i < 2 || n > nums[i-2]) nums[i++] = n; return i; }