LeetCode 2016 273,233,174,149,146

273 Integer to English Words

/*My thoughts1. 写一个专门处理三个数字的函数，注意应对001等Leading zeros的问题2. 空格的处理：当要写入ans时，若ans.size()!=0，加空格后再写入3. 注意单词不要拼写错误4. 有负数*/class Solution {public:    vector<string> Words;    vector<string> conj;    Solution()    {        Words.resize(105);        conj.resize(5);        conj[1]="";        conj[2]="Thousand";        conj[3]="Million";        conj[4]="Billion";        Words[0]="Zero";        Words[1]="One";        Words[2]="Two";        Words[3]="Three";        Words[4]="Four";        Words[5]="Five";        Words[6]="Six";        Words[7]="Seven";        Words[8]="Eight";        Words[9]="Nine";        Words[10]="Ten";        Words[11]="Eleven";        Words[12]="Twelve";        Words[13]="Thirteen";        Words[14]="Fourteen";        Words[15]="Fifteen";        Words[16]="Sixteen";        Words[17]="Seventeen";        Words[18]="Eighteen";        Words[19]="Nineteen";        Words[20]="Twenty";        Words[30]="Thirty";        Words[40]="Forty";        Words[50]="Fifty";        Words[60]="Sixty";        Words[70]="Seventy";        Words[80]="Eighty";        Words[90]="Ninety";        Words[100]="Hundred";    }    string intTostr(int num)    {        stringstream ss;        ss<<num;        string ans = ss.str();        return ans;    }    string Three(string num)    {        string ans = "";        int n = atoi(num.c_str());        if (n>=100 && n<=999)        {            if (n%100==0)            {                ans = Words[num[0]-'0']+" "+Words[100];                return ans;            }            ans = Words[n/100]+" "+Words[100]+" ";            n=n-n/100*100;        }        if(n<=20 || n%10==0)        {            ans+=Words[n]; return ans;        }        ans+=Words[n/10*10]+" "+Words[n%10];        return ans;    }    string numberToWords(int num)    {        string ans ="";        if (num<0) ans="Negative ";        string number = intTostr(num);        int len = number.size();        int cutLen = len - len/3*3;        len = len /3;        if (cutLen!=0) len++; else cutLen = 3;        for(int i=len;i>=1;i--)        {            string tmpNum = number.substr(0,cutLen);            if (tmpNum=="000")            {                if (i!=1)                {                    number=number.substr(cutLen);                }                cutLen = 3;                continue;            }            if (ans.size()!=0) ans+=" ";            ans += Three(tmpNum);            if (i!=1)            {                ans+=" "+conj[i];                number=number.substr(cutLen);            }            cutLen = 3;        }        return ans;    }};// ------------------WA cases-------------/*WA reason: ninety拼写错误Input:1234567891Output:"One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninty One"Expected:"One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"WA reason: 忘记处理所有的000项目Input:1000Output:"One Thousand Zero Hundred"Expected:"One Thousand"WA:没有处理leading zerosInput:1001Output:"One Thousand Zero Hundred One"Expected:"One Thousand One"*/

233 Number of Digit One

// --------------------Seeing Discuss-------------/*url:https://discuss.leetcode.com/topic/18054/4-lines-o-log-n-c-java-pythonGo through the digit positions by using position multiplier m with values 1, 10, 100, 1000, etc.For each position, split the decimal representation into two parts, for example split n=3141592 into a=31415 and b=92 when we're at m=100 for analyzing the hundreds-digit. And then we know that the hundreds-digit of n is 1 for prefixes "" to "3141", i.e., 3142 times. Each of those times is a streak, though. Because it's the hundreds-digit, each streak is 100 long. So (a / 10 + 1) * 100 times, the hundreds-digit is 1.Consider the thousands-digit, i.e., when m=1000. Then a=3141 and b=592. The thousands-digit is 1 for prefixes "" to "314", so 315 times. And each time is a streak of 1000 numbers. However, since the thousands-digit is a 1, the very last streak isn't 1000 numbers but only 593 numbers, for the suffixes "000" to "592". So (a / 10 * 1000) + (b + 1) times, the thousands-digit is 1.The case distincton between the current digit/position being 0, 1 and >=2 can easily be done in one expression. With (a + 8) / 10 you get the number of full streaks, and a % 10 == 1 tells you whether to add a partial streak.*/class Solution {public:    int countDigitOne(int n)    {        int ones = 0;        for(long long m = 1;m<=n;m*=10)        {            int a = n/m, b= n%m;            ones += (a+8)/10*m +(a%10==1)*(b+1);        }        return ones;    }};

174 Dungeon Game

// ---------------------Seeing Discuss--------------/*URL:https://discuss.leetcode.com/topic/6912/c-dp-solutionUse hp[i][j] to store the min hp needed at position (i, j), then do the calculation from right-bottom to left-up.Note: adding dummy row and column would make the code cleaner.Discuss hp[i][j]表示的是，(i,j)位置需要的最少血量；然后倒着循环；并且在计算hp[i][j]时，如果>0 则1，小于0，则血量小结：这种方格dp，算不出来要想着倒着来，或者换个状态！*/class Solution {public:    int calculateMinimumHP(vector<vector<int> >& dungeon)    {        int M = dungeon.size();        int N = dungeon[0].size();        vector< vector<int> >hp(M+1,vector<int>(N+1,INT_MAX));        hp[M][N-1] = 1;        hp[M-1][N] = 1;        for(int i=M-1;i>=0;i--)        {            for(int j=N-1;j>=0;j--)            {                int need = min(hp[i+1][j],hp[i][j+1]) - dungeon[i][j];                hp[i][j]=need<=0? 1:need;            }        }        return hp[0][0];    }};// ---------------------WA---------------/*  35 / 44 test cases passed.WA data pointInput:[[1,-3,3],[0,-2,0],[-3,-3,-3]]Output:5Expected:3找不到最x生命值和当前生命值的平衡。。。class Solution {public:    int calculateMinimumHP(vector<vector<int> >& dungeon)    {        int row = dungeon.size();        if (row==0) return 0;        int col = dungeon[0].size();        vector<vector<int> > f(row+1,vector<int> (col+1,0));        vector<vector<int> > g(row+1,vector<int> (col+1,0));        f[0][0]=dungeon[0][0];        g[0][0]=dungeon[0][0];        for(int i=0;i<row;i++)        {            for(int j=0;j<col;j++)            {                if (i==0 && j==0) continue;                int gTmpUp = INT_MIN, gTmpLeft = INT_MIN;                int fTmpUp=0, fTmpLeft=0;                if (i-1>=0)                {                    fTmpUp = dungeon[i][j]+f[i-1][j];                    gTmpUp = min(g[i-1][j],fTmpUp);                }                if (j-1>=0)                {                    fTmpLeft = dungeon[i][j]+f[i][j-1];                    gTmpLeft = min(g[i][j-1],fTmpLeft);                }                if(gTmpUp > gTmpLeft)                {                    f[i][j]=fTmpUp;                    g[i][j]=gTmpUp;                }                else                {                    f[i][j]=fTmpLeft;                    g[i][j]=gTmpLeft;                }            }        }        if (g[row-1][col-1]>0) return 1;        return abs(g[row-1][col-1])+1;    }};*/

149 Max Points on a Line

// --------------Seeing Discuss---------------/*Url:https://discuss.leetcode.com/topic/6028/sharing-my-simple-solution-with-explanation/2Discuss大致思想：1. 对于每个点p，计算他和其他所有点的斜率，并用map统计每个斜率出现的次数；2. 其中对于与p横纵坐标相同的点，用samePoint来累计3. 横坐标与p同的，map放在INT_MAX里记录4. 每个点算完以后，知道斜率出现次数最多的localmax，来更新全局max5. 因为已经定了点p的位置，也就是这条直线必经过点p，所以，截距也是定的，只需要计算斜率就行了。要点：选定了点p，只计算斜率即可，因为必定过点p，截距也是定的。*/class Solution {public:    int maxPoints(vector<Point>& points)    {        int result = 0;        for(int i=0;i<points.size();i++)        {            int samePoint = 1;            unordered_map<double,int> mapping;            mapping[INT_MAX]=0;            for(int j=i+1;j<points.size();j++)            {                if (points[i].x==points[j].x && points[i].y==points[j].y)                    samePoint++;                else                    if (points[i].x==points[j].x)mapping[INT_MAX]++;                    else                    {                        double slope = double(points[i].y-points[j].y)/double(points[i].x-points[j].x);                        if (mapping.find(slope)==mapping.end())                            mapping[slope]=1;                        else mapping[slope]++;                    }            }            int localMax = 0;            for(auto it = mapping.begin();it!=mapping.end();it++)            {                localMax=max(localMax,it->second);            }            localMax += samePoint;            result = max(result,localMax);        }        return result;    }};/*My Thoughts:1.找到最多的点，使得他们在一条直线上2.先给所有的点，按照x坐标从小到大排序3.两点确定一条直线，斜率，截距可以知道。4.判断其他的点是不是（斜率，截距）组的点即可5.注意跟横纵坐标平行的直线*/

146 LRU Cache

class LRUCache{public:    LRUCache(int capacity)    {        cache.clear();        ranking.clear();        cap = capacity;        pri = 0;    }    int get(int key)    {        if (cache.find(key)!=cache.end())        {            map<int,int>::iterator it = ranking.find(cache[key].first);            ranking.erase(it);            pri++;            cache[key].first = pri;            ranking[pri]=key;            return cache[key].second;        }        return -1;    }    void set(int key, int value)    {        if (cache.find(key)==cache.end() && cache.size()==cap)        {            map<int,int>::iterator it = ranking.begin();            cache.erase(it->second);            ranking.erase(it);        }        if(cache.find(key)!=cache.end())        {            map<int,int>::iterator it = ranking.find(cache[key].first);            ranking.erase(it);        }        pri++;        pair<int,int> x(pri,value);        cache[key]=x;        ranking[pri]=key;        return ;    }private:    map<int,pair<int,int> > cache;// first: the pri; second: the value    map<int,int> ranking; // ranking[pri]= key;    int pri;    int cap;};