【leetcode】格雷编码 背下来即可

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> ans;
        for(int i=0;i<1<<n;i++)
        {
            ans.push_back(i^(i>>1));
        }
        return ans;
    }
};

【复习数组和字符串】

寻找中间的索引

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        if(nums.size()==0)return -1;
        int sum = 0;
        for(int i=0;i<nums.size();i++)
            sum += nums[i];
        
        int leftsum = 0;
        for(int i=0;i<nums.size();i++)
        {
            if(leftsum*2+nums[i] == sum)
                return i;
            leftsum += nums[i];
        }
        return -1;

    }
};

二分寻找插入的位置

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

合并区间

class Solution {
public:
    static bool cmp(vector<int>a,vector<int>b)
    {
        return (a[0]<b[0]);
    }
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int>>ans;
        sort(intervals.begin(),intervals.end(),cmp);

/*
        for(int i=0;i<intervals.size();i++)
        {
            for(int j=0;j<intervals[i].size();j++)
            {
                printf("%d ",intervals[i][j]);
            }
            printf("\n");
        }
*/ 
        if(intervals.size()==0) return ans;
        int left = intervals[0][0];
        int right = intervals[0][1];

        for(int i=0;i<intervals.size();i++)
        {
            if(intervals[i][0]<=right)
            {
                if(right<intervals[i][1])
                    right = intervals[i][1];
            }
            else if(intervals[i][0]>right )
            {
                vector<int> tmp = {left,right};
                ans.push_back(tmp);
                left = intervals[i][0];
                right = intervals[i][1];
            }

            if(i == intervals.size()-1)
            {
                vector<int> tmp = {left,right};
                ans.push_back(tmp);
            }
        }

        return ans;
    }
};

零矩阵

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        if(matrix.size()==0)return;
        int rows[matrix.size()];
        int cols[matrix[0].size()];
        memset(rows,0,sizeof(rows));
        memset(cols,0,sizeof(cols));
        for(int i=0;i<matrix.size();i++)
        {
            for(int j=0;j<matrix[i].size();j++)
            {
                if(matrix[i][j] == 0)
                {
                    rows[i] = 1;
                    cols[j] = 1;
                }
            }
        }
        int m = matrix.size();
        int n = matrix[0].size();
        for(int i=0;i<matrix.size();i++)
        {
            if(rows[i] == 1)
            for(int j=0;j<matrix[i].size();j++)
            {
                    matrix[i][j] = 0;
            }
        }

        for(int j=0;j<n;j++)
        {
            if(cols[j] == 1)
            {
                for(int i=0;i<m;i++)
                {
                    matrix[i][j] = 0;
                }
            }
        }
    }
};

对角线遍历

要注意判断边界条件的优先级

class Solution {
public:
    vector<int> findDiagonalOrder(vector<vector<int>>& matrix) {
        vector<int> ans;

        int m = matrix.size();
        if(m==0)return ans;
        int n = matrix[0].size();
        int ccount = m*n;
        int x = 0;
        int y = 0;
        while(ccount--)
        {
            ans.push_back(matrix[x][y]);
            printf("%d %d\n",x,y);

            if(x == m-1&&((x+y)%2==1))
            {
                y++;
            }
            else if((y ==0&&(x+y)%2==1))
            {
                x++;
            }
            else if(y == n-1&&(x+y)%2==0)
            {
                x++;
            }
            else if(x == 0&&((x+y)%2==0))
            {
                y++;
            }
            else if((x+y)%2==0)
            {
                x--;
                y++;
            }
            else if((x+y)%2==1)
            {
                x++;
                y--;
            }
        }
        return ans;
    }
};

字符串最长公共前缀

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        string ans;
        if(strs.size() == 0) return "";

        for(int j=0;j<strs[0].length();j++)
        {
            char tmp = strs[0][j];
            int is = 1;
            for(int i=0;i<strs.size();i++)
            {
                if(strs[i][j]!=tmp)
                {
                    is = 0;
                    break;
                }
            }
            if(is)
            ans.push_back(tmp);
            else break;
        }
        return ans;
    }
};

 最长回文子串

动态规划

class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.length();
        if(n<=1)return s;
        string ans;
        int dp[n][n];
        int maxlen = -1;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<=i;j++)
            {
                if(s[j] == s[i])
                {
                    if(i-j>=3)
                    {
                        dp[j][i] = dp[j+1][i-1];
                    }
                    else dp[j][i] = 1;
                }
                else 
                {
                    dp[j][i] = 0;
                }

                if(dp[j][i] == 1&&i-j+1>maxlen)
                {
                    maxlen = i-j+1;
                    ans = s.substr(j,maxlen);
                    printf("%d %d\n",j,i);
                }
                
            }
                    

        }
        return ans;


    }
};

翻转句子里的单词

要求使用空间为o(1)

使用stl的reverse函数

class Solution {
public:
    string reverseWords(string s) {
        reverse(s.begin(),s.end());

        int left,right;
        for(int i=0;i<s.length();i++)
        {
            if(s[i] == ' ') continue;
            if(isalpha(s[i])||isdigit(s[i]))
            {
                left = i;
                i++;
                while(isalpha(s[i])||isdigit(s[i]))
                {
                    i++;
                }
                right = i;

                reverse(&s[left],&s[right]);
            }
        }

        int pos = 0;
        int word = 0;
        for(int i=0;i<s.length();i++)
        {
            if((s[i] == ' '&&pos == 0)||(s[i] == ' '&&word == 0))
            {
                continue;
            }
            else if(isalpha(s[i])||isdigit(s[i]))
            {
                word = 1;
                s[pos++] = s[i];
                continue;
            }
            else if(s[i] == ' '&&word ==1)
            {
                s[pos++] = s[i];
                word = 0;
            }

        }

        if(s[pos-1]==' ')
            pos--;
        s = s.substr(0,pos);
        return s;
        
    }
};