算法学习笔记（LeetCode OJ）

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LeetCode的一些算法题，都是自己做的，欢迎提出改进~~

LeetCode：http://oj.leetcode.com

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<Reverse Words in a String>-20140328

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.

 public class Solution {
     public String reverseWords(String s) {
         s = s.trim();
         String[] str = s.split("\\s{1,}");
         String tmp;
         int len = str.length;
         int halfLen = len/2;
         for(int i=0;i<halfLen;i++){
             tmp = str[i];
             str[i] = str[len-i-1];
             str[len-i-1] = tmp;
         }
         StringBuffer sb = new StringBuffer();
         String result;
         if(len==1){
             result = str[0];
         }else{
             for(String string:str){
                 sb.append(string+" ");
             }
             result = sb.toString().trim();
         }
         return result;
     }
 }

Java Code - 404ms - AC1/7

C++: http://blog.csdn.net/feliciafay/article/details/20884583

一个空格的情况
多个空格在一起的情况
空串的情况
首尾空格的情况
一些测试数据：
“            ”
“a       ”
"       a"
"      a     "
"     a      b    "
"     a b     "
"     a  b"
"a  b       "

Hint

<Evaluate Reverse Polish Notation>-20140329

Evaluate the value of an arithmetic expression in Reverse Polish Notation.（逆波兰表达式/后缀表达式）
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:
  ["", "", "+", "", "*"] -> (( + ) * ) ->
  ["", "", "", "/", "+"] -> ( + ( / )) -> 

 public class Solution {
     public int evalRPN(String[] tokens) {
         Stack<Integer> stack = new Stack();
         for(String str:tokens){
             if("+-*/".contains(str)){
                 int b = stack.pop();
                 int a = stack.pop();
                 switch(str){
                     case "+":
                         stack.add(a+b);
                         break;
                     case "-":
                         stack.add(a-b);
                         break;
                     case "*":
                         stack.add(a*b);
                         break;
                     case "/":
                         stack.add(a/b);
                         break;
                 }
             }else{
                 stack.add(Integer.parseInt(str));
             }
         }
         return stack.pop();
     }
 }

Java Code - 484ms - AC1/1

 <Max Points on a Line>-20140330

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

开始的实现：

 /**
  * Definition for a point.
  * class Point {
  *     int x;
  *     int y;
  *     Point() { x = 0; y = 0; }
  *     Point(int a, int b) { x = a; y = b; }
  * }
  */
 public class Solution {
     private boolean judgment(Point a, Point b, Point third){
         return (a.y-third.y)*(b.x-third.x)==(b.y-third.y)*(a.x-third.x);
     }

     public int maxPoints(Point[] points) {
         int len = points.length;
         if(len<3){
             return len;
         }
         int[] mark = new int[len];
         for(int m=0;m<len;m++){
             mark[m] = 0;
         }
         int count = 2;
         for(int i=0;i<len;i++){
             int tmpCount = 2;
             for(int j=i+1;j<len;j++){
                 if(mark[j]==1){
                     continue;
                 }
                 for(int k=j+1;k<len;k++){
                     if(mark[k]==1){
                         continue;
                     }
                     boolean judge = judgment(points[i],points[j],points[k]);
                     if(judge){
                         mark[k] = 1;
                         tmpCount++;
                     }
                 }
             }
             if(tmpCount>count){
                 count = tmpCount;
             }
         }
         return count;
     }
 }

Java Code - Wrong

后来发现被坑了，它是有相同点出现这种情况的！！！开始我还傻傻地想为什么<5,6,18>三点共线而<0,5,6>却不共线，后来自己费了好大力气才明白过来。

然后，还有其他的问题，对此题无力了，还是先放放过几天头脑清醒再来弄吧T-T。

看看人家的思想先吧，不过我还是想先自己搞掂再看。

C++: http://blog.csdn.net/feliciafay/article/details/20704629

<Single Number>-20140414

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

 public class Solution {
     public int singleNumber(int[] A) {
         int num = 0;
         for(int i : A){
             num ^= i;
         }
         return num;
     }
 }

Java Code