Remove Duplicates from Sorted Array

Easy 原题链接：https://leetcode.com/problems/remove-duplicates-from-sorted-array/

Remove Duplicates from Sorted Array

原题链接：https://leetcode.com/problems/remove-duplicates-from-sorted-array/

描述

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.

例子

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.

• 注意

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.

解法一

使用index，在原数组上做remove

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

或者

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

分析：

T=O(n), S=O(1)

解法二

使用STL的distance和unique

关于distance：

http://cplusplus.com/reference/iterator/distance/?kw=distance

Calculates the number of elements between first and last.

关于unique：

http://cplusplus.com/reference/algorithm/unique/?kw=unique

Removes all but the first element from every consecutive group of equivalent elements in the range [first,last).

class Solution {
    public:
    int removeDuplicates(vector<int>& nums) {
        return distance(nums.begin(), unique(nums.begin(), nums.end()));
    } 
};

分析：

根据STL中unique是使用双指针实现的：

template <class ForwardIterator>
  ForwardIterator unique (ForwardIterator first, ForwardIterator last)
{
  if (first==last) return last;

  ForwardIterator result = first;
  while (++first != last)
  {
    if (!(*result == *first))  // or: if (!pred(*result,*first)) for version (2)
      *(++result)=*first;
  }
  return ++result;
}

解法三

改写unique，使用upper_bound

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        return distance(nums.begin(), removeDuplicates(nums.begin(),nums.end(), nums.begin()));
    }
    template<typename InIt, typename OutIt>OutIt removeDuplicates(InIt first, InIt last, OutIt output) {
        while (first != last) {
            *output++ = *first;
            first = upper_bound(first, last, *first);
        }
        return output;
    }
};

分析：

总结

题目开始，联想到快慢指针的方法。如果对STL熟悉，可以想到unique函数来进行解题，但是题目提示为sorted array，所以unique的解法不一定是最优解。在理解unique的原理和分析题目的基础上，作出变形改写。