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80. Remove Duplicates from Sorted Array II

80. Remove Duplicates from Sorted Array II

题目

 Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array 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. It doesn't matter what you leave beyond the new length. 

解析

方法一:很灵活的方法,扩展性较强,如果将occur<2 改为   occur<3  ,就变成了允许重复最多三次
	
class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if(n<=2) return n;
        int index=2;//允许重复两次,可以修改为三次
        for(int i=2;i<n;i++)
        {
            if(A[i]!=A[index-2])//允许重复两次,可以修改为三次
                A[index++]=A[i];
        }
         
        return index;
    }
};

方法二:简洁版本

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        int index=0;
        for(int i=0;i<n;i++){
            if(i>0&&i<n-1&&A[i]==A[i-1]&&A[i]==A[i+1])
                continue;
             
            A[index++]=A[i];
        }
         
        return index;
    }
};
// 80. Remove Duplicates from Sorted Array II
class Solution_80 {
public:
	int removeDuplicates_(vector<int>& nums) {

		if (nums.size()<=1)
		{
			return nums.size();
		}

		int len = 0;
		int start = 0, end = 0;

		for (int i = 1; i < nums.size();i++)
		{
			if (nums[i]==nums[i-1])
			{
				end++;
			}
			else
			{
				start = i;
				end = i;
			}
			if (end-start+1<=2)
			{
				nums[++len] = nums[i];
			}
		}
		
		return len+1;
	}

	int removeDuplicates(int A[], int n) {
		
		vector<int> vec(A, A + n); //vec传值不能达到A;
		return removeDuplicates_(vec);
	}

	int removeDuplicates_1(int A[], int n) {
		int *nums = A;
		if (n <= 1)
		{
			return n;
		}

		int len = 0;
		int start = 0, end = 0;

		for (int i = 1; i < n; i++)
		{
			if (nums[i] == nums[i - 1])
			{
				end++;
			}
			else
			{
				start = i;
				end = i;
			}
			if (end - start + 1 <= 2)
			{
				nums[++len] = nums[i];
			}
		}

		return len + 1;
	}


};

题目来源

posted @ 2018-04-04 15:55  ranjiewen  阅读(156)  评论(0编辑  收藏  举报