Remove Duplicates from Sorted Array

每日算法——letcode系列

问题 Remove Duplicates from Sorted Array

Difficulty: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.

// C++
class Solution {
public:
   int removeDuplicates(vector<int>& nums) {
       
   }
};

翻译 移除有序数组中重复元素

难度系数：容易

给定一个有序的数组，删除数组的重复的元素使得每个元素只出现一 次，并返回数组的长度

你只能在常量内存空间中操作，不能再额外分配一个数组空间

举例：
假定数组 nums = [1,1,2],

函数返回的长度应 = 2, 数组的前 两个数 分别应为1和2

思路

方案一： STL中的unique和distance
方案二： 维护一个index记录不重复的元素

代码

方案一：

// C++
class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        auto firstIterator = nums.begin();
        auto endIterator = nums.end();
        return static_cast<int>(distance(firstIterator, unique(firstIterator, endIterator)));
    }
};

方案二：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()){
            return 0;
        }
        
        int index = 0;
        for (size_t i = 1; i < nums.size(); ++i) {
            if (nums[index] != nums[i]){
                nums[++index] = nums[i];
            }
        }
        return ++index;
    }
};

延伸思考

删除让我想到一个出自《编程珠玑》的找重复的算法题

英文版：

Given a sequential file containing 4,300,000,000 32-bit intergers, how can you find one that appears at least twice?

中文版：

给定包含4 300 000 000 个 32 位整数的顺序文件， 如何找出一个出现至少两次的整数？