26. Remove Duplicates from Sorted Array【easy】

26. Remove Duplicates from Sorted Array【easy】

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.

解法一：

 1 class Solution {
 2 public:
 3     int removeDuplicates(vector<int>& nums) {
 4         if (nums.empty() || nums.size() == 1)
 5         {
 6             return nums.size();
 7         }
 8
 9         int i = 0;
10         int j = 0;
11         nums[j++] = nums[i++];
12
13         while (i < nums.size() && j < nums.size())
14         {
15             if (nums[i] != nums[i - 1])
16             {
17                 nums[j++] = nums[i++];
18             }
19             else
20             {
21                 ++i;
22             }
23         }
24
25         return j;
26     }
27 };

思路：双指针，注意边界条件的判断

解法二：

 1 class Solution {
 2     public:
 3     int removeDuplicates(int A[], int n) {
 4         if(n < 2) return n;
 5         int id = 1;
 6         for(int i = 1; i < n; ++i)
 7             if(A[i] != A[i-1]) A[id++] = A[i];
 8         return id;
 9     }
10 };

可读性一般

解法三：

 1 class Solution {
 2 public:
 3     int removeDuplicates(vector<int>& nums) {
 4         int count = 0;
 5         for(int i = 1; i < nums.size(); i++){
 6             if(nums[i] == nums[i-1]) count++;
 7             else nums[i-count] = nums[i];
 8         }
 9         return nums.size()-count;
10     }
11 };

转载于:https://www.cnblogs.com/abc-begin/p/7538288.html

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