Leetcode[26]-Remove Duplicates from Sorted Array

Link: https://leetcode.com/problems/remove-duplicates-from-sorted-array/

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.

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思路：须要额外加入一个新的变量pos。初始值为0。用来记录非反复元素的位置。从数组的第二个元素開始遍历，假设和前面的元素相等，则直接跳到下一个。假设不等，则将该数组的值赋值给++pos位，接着继续遍历下一个；

Code(c++):

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

做个简单的改动：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int n = nums.size();
        if(n==0) return 0;
        int pos = 1,i = 1;//将pos从1開始
        while(i < n){
            if(nums[i] == nums[i-1]){
                i++;
            }else{
                nums[pos++] = nums[i++];//这里保持一致
            }
        }
        n = pos;
        nums.resize(n);
        return n;
    }
};

Python代码

class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        if n == 0: 
            return 0
        pos = 1;i = 1;count=0
        while i < n:
            if nums[i] == nums[i-1]:
                i=i+1
                count=count+1
            else:
                nums[pos] = nums[i]
                pos = pos+1
                i = i + 1
        for i in range(count):
            nums.pop(-1)
        return len(nums)