Leetcode刷题127-80. 删除排序数组中的重复项 II(C++最简单实现详细解法！！！)

Come from : [https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array-ii/submissions/]

1.Question

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice 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 = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2 :

Given nums = [0,0,1,1,1,1,2,3,3],

Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 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]);
}

2.Answer

medium类型题目。。。。

AC代码如下：（暴力法）

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int index = 2;
        if(nums.size() < 3)
        {
            return nums.size();
        }
        for(int i = index; i < nums.size(); i++) 
        {
    	    if(nums[index-2] != nums[i]) 
            {
    		    nums[index++] = nums[i];
    	    }
        }
    return index;
    }
};

3.大神解答

AC代码如下：

class Solution {
public:
	int removeDuplicates(vector<int>& nums) {
		unordered_map<int, int>	maps;
		int n = 0;
		for (int i = 0; i < nums.size(); ++i) {
			if (maps[nums[i]] < 2) {
				maps[nums[i]]++;
				n++;
			}
			else {
				nums.erase(nums.begin() + i);
				--i;
			}
				
		}
		return n;
	}
};

4.我的收获

fighting。。。
明天回家喽。。。开心，，，
在家呆个十天左右吧~

2019/7/19 胡云层 于南京 127