leetcode 26. Remove duplicates in ordered array

You are given an   array arranged in  ascending ordernums  . Please  delete the repeated elements   in place so that each element appears only once  and return the new length of the array after deletion. The relative order of elements   should remain  consistent  . Then return  nums the number of unique elements in .

Considering  nums the number of unique elements is  k , you need to do the following to ensure that your solution can be passed:

• Changes the array  nums so that  nums the previous  k element of contains unique elements,  nums arranged in the order in which they originally appeared in . nums The size of the remaining elements  nums does not matter.
• return k 

26. Remove duplicates in ordered array - LeetCode

My solution to the problem:

// 推导过程 （双指针）
  |                
0,0,1,1,1,2,2,3,3,4
|

    |                
0,0,1,1,1,2,2,3,3,4
|

    |                
0,1,1,1,1,2,2,3,3,4
  |

          |                
0,1,1,1,1,2,2,3,3,4
  |

          |                
0,1,1,1,1,2,2,3,3,4
    |

              |                
0,1,2,1,1,2,2,3,3,4
    |

              |                
0,1,2,1,1,2,2,3,3,4
      |


              |                
0,1,2,3,1,2,2,3,3,4
      |
    
                  |                
0,1,2,3,1,2,2,3,3,4
      |

                  |                
0,1,2,3,1,2,2,3,3,4
        |

                    |                
0,1,2,3,4,2,2,3,3,4
        |

My code:

// 26. 删除有序数组中的重复项 (请你 原地 删除重复出现的元素)
class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int i=0,j=1;
        if(nums.size()==0) return 0;
        if(nums.size() == 1) return 1;
        for(;i<nums.size() && j<nums.size();) {
            while(j < nums.size() && (nums[i] == nums[j])) {
                j++;
            }
            ++i;
            if(i<nums.size() && j<nums.size())
                nums[i] = nums[j];
        }
        return i;
    }
};

Simplify code

// 简化
class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.size()==0) return 0;
        int p=0,q=1;
        while(q < nums.size()) {
            if(nums[p]!=nums[q]) {
                nums[p+1] = nums[q];
                p++;
            }
            q++;
        }
        return p+1;
    }
};

Official code:

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

作者：力扣官方题解
链接：https://leetcode.cn/problems/remove-duplicates-from-sorted-array/solutions/728105/shan-chu-pai-xu-shu-zu-zhong-de-zhong-fu-tudo/
来源：力扣（LeetCode）
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Complexity analysis

Time complexity: O(n)O(n)O(n), where nnn is the length of the array. The fast pointer and slow pointer can each move at most nnn times.

Space complexity: O(1)O(1)O(1). Just need to use the extra space of the constant