Leetcode初级算法（初更）

Leetcode (初级算法)

26.删除排序数组中的重复项：

时间复杂度：O(n)
空间复杂度：O(1)

C++:

class Solution {
public:
    int removeDuplicates(vector& nums) {
        if(nums.size()==0)return 0;
            int i=0;
            for(vector::iterator itr= nums.begin()+1; itr!= nums.end() ;itr++)
            {
                if (nums[i]!=*itr)
                {
                    i++;
                    nums[i]=*itr;
                }

            }
        return i+1;
    }
};

26ms->96.44%

C:

int removeDuplicates(int* nums, int numsSize) {
    if(numsSize==0)return 0;
    int i=0;
    for(int j=1;jif(nums[i]!=nums[j])
        {
            i++;
            nums[i]=nums[j];
        }
    }
    return i+1;
}

16ms->59.21%

int removeDuplicates(int* nums, int numsSize) {
    if (numsSize == 0)return 0;
    int *i = nums;
    int c = 0;
    for (int j=0; j1; j++)
    {
        nums++;
        if (*i != *nums)
        {
            i++;
            c++;
            *i =*nums;
        }
    }
    return c + 1;
}

12ms->87.59%

659. Split Array into Consecutive Subsequences：

C++(1):

bool isPossible(vector<int>& nums) {
    int i = 0;
    int cut = 0;
    for (int j = 0; j1; j++)
    {
        if (nums[j + 1] == nums[j] + 1)
        {
            i++;
        }
        else if (nums[j + 1] == nums[j])
        {
            int re = 2;
            for (int k = j + 2; k < nums.size(); k++)
            {
                if (k + re - 1 > nums.size() - 1)
                    break;
                if (nums[k] == nums[j])
                {
                    re++;

                }
                else if (nums[k+re-1] == nums[j] + 1)
                {
                    nums.insert(nums.begin() + j+1, nums[k]);
                    nums.erase(nums.begin() + k+1);         
                    i++;
                    j++;
                    re=1;
                }
                else if (nums[k + re-1] > (nums[j] + 1))
                {
                    break;
                }
            }
            cut =cut+ i+1;
            //j = cut-1;
            if (i < 2)
            {
                return false;
            }
            else
            {
                i = 0;
            }
        }
        else
        {
            if (i < 2)
            {
                return false;
            }
            else
            {
                cut = cut + i + 1;
                j = cut;
                i = 0;
            }
        }
    }
    if (i < 2)
    {
        return false;
    }
    return true;
    }

124ms

    bool isPossible(vector<int>& nums) {
    unordered_map<int,int> a, b;
    for(vector<int>::iterator i=nums.begin();i!=nums.end();i++)
        a[*i]++;
    for(vector<int>::iterator i=nums.begin();i!=nums.end();i++)
    {
        if(!a[*i]) continue;
        if(b[*i]>0)
          a[*i]--,b[*i]--,b[*i+1]++;
        else if(a[*i+1]&&a[*i+2])
          a[*i]--,a[*i+1]--,a[*i+2]--,b[*i+3]++;
        else return false;
    }
        return true;
    }

92ms->79.31%

258. 各位相加：

java:

    public int addDigits(int num) {
        int tmp=0;
        while(num>9)
        {
            tmp=0;
            while(num>0)
            {
                tmp=tmp+num%10;
                num=num/10;
            }
            num=tmp;
        }
        return num;
    }

4ms

    public int addDigits3(int num) {
        if (num < 10 || num==0) {
            return num;
        }
        num=num%9;
        if(num==0)
            num = 9;
        return num;
    }

2ms