leetcode day1 -- Reverse Words in a String Evaluate Reverse Polish Notation Max Points on a Li

以前从来没做过什么oj，发现做oj和在本地写代码或者纸上写差别还是很大的，觉得今天开始刷oj,特此记录一下。

1、Reverse Words in a String

Given an input string, reverse the string word by word.

For example,
Given s = "the sky is blue",
return "blue is sky the".

click to show clarification.

Clarification:

What constitutes a word?
A sequence of non-space characters constitutes a word.
Could the input string contain leading or trailing spaces?
Yes. However, your reversed string should not contain leading or trailing spaces.
How about multiple spaces between two words?
Reduce them to a single space in the reversed string.

此题在于看清题意，要去掉头尾的空格，单词中间多空格只保留一个，注意string类型的使用方法，查的http://www.cplusplus.com/reference/string/string/

去掉多连续空格时使用了stringstream。

运行成功的代码如下：

class Solution {
public:void reverseWords(string &s) {if(s.empty()){return;}int start = 0;int end = s.length()-1;reverseWord(s,start,end);removeSpaces(s);int length = s.length();for(int i=0; i <= length; ++i){if(i == length || s.at(i) == ' ' && i>=1){end = i-1;reverseWord(s,start,end);start = end = i+1;}}}
private:void reverseWord(string &s,int start, int end){while( start < end ){char temp = s.at(start);s.at(start) = s.at(end);s.at(end) = temp;++start;--end;}}void removeSpaces(string& s){stringstream ss;ss << s;string t;s="";while(ss >> t){s += t+" ";}if(s.length()>1){s.erase(s.end()-1);}}
};

2、Evaluate Reverse Polish Notation

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Some examples:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

代码如下：用栈来实现：

注意易错点：（1）除0问题（2）输入一个数字时输出问题（3）错误处理问题，如果输入出错时输出结果为多少

class Solution {
public:int evalRPN(vector<string> &tokens) {int result = 0;for(int i=0; i<tokens.size(); ++i){int temp = operatorType(tokens.at(i));if( temp == NUMBER){result = atoi(tokens.at(i).c_str());dataStack.push(result);}else {if(dataStack.size()<2){return result;}int param2 = dataStack.top();dataStack.pop();int param1  = dataStack.top();dataStack.pop();switch(temp){case ADD:result = param1 + param2;break;case MINUS:result = param1 - param2;break;case MULTIPLY:result = param1 * param2;break;case DIV:if(param2 == 0){return 0;}result = param1 / param2;break;}dataStack.push(result);}}return result;}
private:stack<int> dataStack;enum operat{NUMBER,ADD,MINUS,MULTIPLY,DIV};int operatorType(string& s){if(s.length()>1){return NUMBER;}else{switch(s.at(0)){case '+':return ADD;case '-':return MINUS;case '*':return MULTIPLY;case '/':return DIV;default:return NUMBER;}}}
};

3、Max Points on a Line

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

分析：对每个点求和其他点的夹角的正切值和数量，建立map

易错点：（1）特殊情况下，点的个数为0，1，,2时（2）相同的点，肯定在一条直线上；（3）横坐标相同的点，正切值无穷大，单独计算

在下面的代码中，两个点的正切值计算了两次，其实可以计算一次，但是实现比较复杂，如果用map来存储的话，存储point为键值时，需要自动排序，需要Point < 的定义，还是采用了下面的方法：

代码：

class Solution {
public:int maxPoints(vector<Point> &points) {if( points.size() <= 2){return points.size();}map<float,int> tempHash;float tanAngle = 0.0;int maxPoints = 2;map<float,int>::iterator iter;for( int i=0; i< points.size(); ++i){tempHash.clear();int vertiLineNum = 1; //垂直线点数int samePointNum = 0; //坐标相同的点数for( int j=0; j<points.size(); ++j){if( i == j){ //同一个点忽略continue;}//如果x坐标相同if(points.at(i).x == points.at(j).x){//如果x，y坐标相同,则为同一个点if(points.at(i).y == points.at(j).y){++ samePointNum;continue;}//否则，位于同一垂直线上++ vertiLineNum;}else{//根据倾斜角的正切值来建立哈希表，正切值为键值，值为点数tanAngle = (float)(points.at(j).y-points.at(i).y)/(points.at(j).x-points.at(i).x);if(tempHash[tanAngle]){++tempHash[tanAngle];}else{tempHash[tanAngle] = 2;}}}//更新最大点数maxPoints = max(vertiLineNum + samePointNum,maxPoints);for(iter = tempHash.begin(); iter!=tempHash.end(); ++iter){if((*iter).second + samePointNum > maxPoints){maxPoints = (*iter).second + samePointNum;}}  }return maxPoints;}
};

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