26. Remove Duplicates from Sorted Array[E]刪除排序數組中的重複項

題目

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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.
Example 2：
 Given nums = [0,0,1,1,1,2,2,3,3,4],
 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.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well. Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy) 
int len = removeDuplicates(nums); 
// any modification to nums in your function would be known by the caller. 
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) { 
    print(nums[i]); 
}

思路

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雙指針法

本題要返回一個有序數組內不重複元素的個數，要比較數組內兩個元素是否相同，須要兩個下標記錄，因而想到雙指針法。這裏要注意兩點：一是函數的形參是傳引用，改變形參也會改變本來的數據；二是題目要求不得聲明額外的空間。
對於一個有序數組，因爲不能使用額外的空間保存刪除重複元素後的數組，因而想到不刪除重複元素，而是替換重複元素，將重複元素替換到後面，使用兩個指針：

• pBegin：記錄不重複數組的最後一個元素的下標。
  • 遇到不重複元素時，pBegin加1，表示多了一個不重複元素。
• pEnd：用於循環。
  • 當遇到重複元素時，pEnd加1
  • 當遇到不重複元素時，先pBegin加1，此時pBegin指向的仍是重複元素，pEnd指向不重複元素。將pBegin所指元素與pEnd所指元素替換，此時pBegin指向不重複元素，而且下標表明不重複數組的最後一個下標；pEnd指向重複元素。

C++

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        
       if(nums.size() == 0)
           return 0;
        
        int pBegin = 0;
        int pEnd = 1; 
       
        while(pEnd < nums.size()){
             if(nums[pEnd] != nums[pBegin]){
                pBegin ++;
                nums[pBegin] = nums[pEnd];
            }
            pEnd ++;
        }
        return pBegin + 1;   //從0計數的，所以長度+1
    }
};

Python