26. Remove Duplicates from Sorted Array

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：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int i=0, j=1;
        int n=nums.size();
        if(n==0) return 0;
        while(j<n) {
            if(nums[i]!=nums[j]) {
                i++;
                swap(nums[i], nums[j]);
            }
            j++;
        }
        return i+1;
    }
};