问题：
去除有序数组中重复的数字。
Example 1: Input: nums = [1,1,2] Output: 2, nums = [1,2] Explanation: 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. Example 2: Input: nums = [0,0,1,1,1,2,2,3,3,4] Output: 5, nums = [0,1,2,3,4] Explanation: Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length. Constraints: 0 <= nums.length <= 3 * 10^4 -10^4 <= nums[i] <= 10^4 nums is sorted in ascending order.
解法：slow-fast pointers（快慢指针法）
- 0～low：代表不重复的数组。
- fast：去检测重复的元素，若找到一个不重复的数，
- swap(low+1, fast)
- low++
将不重复的元素 fast 交换至 前面第一个不符合要求的位置 low+1
继续探测下一个元素 fast++
代码参考：
1 class Solution { 2 public: 3 int removeDuplicates(vector<int>& nums) { 4 int i=0, j=1; 5 int n=nums.size(); 6 if(n==0) return 0; 7 while(j<n) { 8 if(nums[i]!=nums[j]) { 9 i++; 10 swap(nums[i], nums[j]); 11 } 12 j++; 13 } 14 return i+1; 15 } 16 };