算法 day6

【leetcode】環形鏈表2

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode* fast  = head;
        ListNode* slow = head;

        while(fast!=NULL && slow!=NULL)
        {
            fast = fast->next;
            slow = slow->next;
            if(fast!=NULL)fast = fast->next;
            else return NULL;
            if(fast == slow)
            break;
        }

        if(fast!=NULL&&slow!=NULL&&fast == slow)
        {
            fast = head;
            while(fast!=slow)
            {
                fast = fast->next;
                slow = slow->next;
            }
            return fast;
        }
        else return NULL;
    }
};

繼續複習數組、字符串。

實現strstr()

使用KMP算法 時間複雜度O(m+n)

class Solution {
public:
    void getnext(int* next,string p)
    {
        int j = -1;
        next[0] = j;
        for(int i=1;i<p.length();i++)
        {
            while(j>=0 && p[j+1]!=p[i])
            {
                j = next[j];
            }
            if(p[j+1]==p[i])
                j++;
            next[i] = j;
        }
    }
    int strStr(string s, string p) {
        if(s.length() == 0)
        {
            if(p.length()!=0)
            return -1;
            else return 0;
        }

        if(p.length()==0)return 0;
        int next[p.length()];

        getnext(next,p);

        int j=-1;

        for(int i=0;i<s.length();i++)
        {
            while(j>=0 && s[i]!=p[j+1])
            {
                j = next[j];
            }

            if(s[i] == p[j+1])
            {
                j++;
            }

            if(j == p.length()-1)
            {
                return i-p.length()+1;
            }
        }
        return -1;
    }
};

數組拆分

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int n = nums.size()/2;
        int sum = 0;
        for(int i=0;i<n;i++)
        {
            sum+=nums[i*2];
        }
        return sum;
    }
};

兩數之和

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int left = 0;
        int right = numbers.size()-1;
        int sum = 0;
        vector<int> ans;
        while(left<right)
        {
            sum = numbers[left]+numbers[right];
            if(sum>target)
            {
                right--;
            }
            else if(sum<target)
            {
                left++;
            }
            else 
            {
                ans.push_back(left+1);
                ans.push_back(right+1);
                return ans;
            }

        }
        return ans;
    }
};

移除指定元素o(n)

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int ccount = 0;
        for(int i=0;i<nums.size();i++)
        {
            if(nums[i]!=val)
            {
                nums[ccount++] = nums[i];
            }
        }
        return ccount;
    }
};

連續最大的1

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int ccount = 0;
        int max = -1;
        for(int i=0;i<nums.size();i++)
        {
            if(nums[i] == 1)
                ccount++;
            if(nums[i] == 0 || i == nums.size()-1)
            {
                if(ccount>max)max = ccount;
                ccount = 0;
            }
        }
        return max;
    }
};

連續長度最小數組

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int n = nums.size();
        if(n==0)return 0;
        int left = 0;
        int right = 1;
        int sum = nums[0];
        int len = INT_MAX;
        
        //[a,b) 左閉右開
        while(left<n&&right<n+1)
        {
            if(sum<s)
            {
                if(right<n)
                sum+=nums[right];
                right++;
            }
            else if(sum>=s)
            {
                len = min(len,right-left);
                sum-=nums[left];
                left++;
            }
            printf("%d %d %d\n",left,right,sum);
        }
        if(len == INT_MAX)return 0;
        return len;
    }
};

楊輝三角

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> ans;
        vector<int> row;
        if(numRows == 0) return ans;
        row.push_back(1);
        ans.push_back(row);
        for(int i=1;i<numRows;i++)
        {
            vector<int> row;
            for(int j=0;j<i+1;j++)
            {
                if(j == 0)row.push_back(1);
                else if(j == i)row.push_back(1);
                else 
                {
                    row.push_back(ans[i-1][j-1]+ans[i-1][j]);
                }

            }
            ans.push_back(row);
        }
        return ans;
    }
};

 楊輝三角2

注意組合數的算法c(n,k+1) = (n-k)/k+1 * c(n,k)

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> ans;
        ans.push_back(1);
        if(rowIndex == 0)return ans;
        long long pre = 1;
        for(int i=1;i<rowIndex;i++)
        {
            long long tmp = pre*(rowIndex-i+1)/i;

            ans.push_back(tmp);
            pre = tmp;
        }
        ans.push_back(1);
        return ans;
    }
};

翻轉字母

class Solution {
public:
    string reverseWords(string s) {
        int left = 0;
        int word = 0;

        for(int i=0;i<s.length();i++)
        {
            if(s[i]!=' ')
            {
                if(word == 0 )
                {
                    printf("%d\n",i);
                    left = i;
                    word = 1;
                }
            }
            if(i == s.length()-1||s[i+1] == ' ' )
            {
                if(word == 1)
                {
                    reverse(&s[left],&s[i+1]);
                    word = 0;
                }
            }
        }
        return s;
    }
};

旋轉數組找最小

class Solution {
public:
    int findMin(vector<int>& a) {
        for(int i=0;i<a.size()-1;i++)
        {
            if(a[i]>a[i+1])
            {
                return a[i+1];
            }
        }
        return a[0];
    }
};

其實是一個二分問題，要使用二分法

class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0;
        int right = nums.size()-1;
        while(left<right)
        {
            int mid = (right-left)/2 + left;
            if(nums[mid]<=nums[right])
            {
                right = mid;
            }
            else if(nums[mid]>nums[right]) 
            {
                left = mid+1;
            }
        }
        return nums[left];
    }
};

刪除重複的元素

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

把所有的0移動到數列的最末端，時間複雜度o(n)

class Solution {
public:

    void moveZeroes(vector<int>& nums) {
        int ccount = 0;
        int left,right;
        left = right = 0;
        while(right<nums.size())
        {
            if(nums[right]==0)
            {
                right++;
                ccount++;
            }
            else
            {
                nums[left] = nums[right];
                right++;
                left++;
            }
        }
        while(ccount--)
        {
            nums[left++] = 0;
        }

    }
};

數組字符串練習完結~~撒花*★,°*:.☆(￣▽￣)/$:*.°★* 。