Description

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]); }

解題思路

這個題意很簡單，就是求出陣列中有幾個不相同的元素。唯一的要求就是不能新建一個數組，要保證O(1)的空間複雜度。

既然不能新建陣列，那麼就設定一個計數器index。

然後遍歷一遍陣列，遍歷過程中，比較nums[index]和nums[i]是否相同，若不相同，就把新的元素放在nums[index]位置上，最後輸出index的大小就好了。

程式碼

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
    	//判斷是否為空陣列
        if(nums.empty()) return 0;
        int index=0;
        for(int i=1;i<nums.size();i++)
        {
            if(nums[index]!=nums[i])
                nums[++index]=nums[i];
        }
        return index+1;
    }
};