算法学习笔记（LeetCode OJ）

==================================

LeetCode的一些算法题，都是自己做的，欢迎提出改进~~

LeetCode：http://oj.leetcode.com

==================================

<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.

 1 public class Solution {

 2     public String reverseWords(String s) {

 3         s = s.trim();

 4         String[] str = s.split("\\s{1,}");

 5         String tmp;

 6         int len = str.length;

 7         int halfLen = len/2;

 8         for(int i=0;i<halfLen;i++){

 9             tmp = str[i];

10             str[i] = str[len-i-1];

11             str[len-i-1] = tmp;

12         }

13         StringBuffer sb = new StringBuffer();

14         String result;

15         if(len==1){

16             result = str[0];

17         }else{

18             for(String string:str){

19                 sb.append(string+" ");

20             }

21             result = sb.toString().trim();

22         }

23         return result;

24     }

25 }

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:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9

  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

 1 public class Solution {

 2     public int evalRPN(String[] tokens) {

 3         Stack<Integer> stack = new Stack();

 4         for(String str:tokens){

 5             if("+-*/".contains(str)){

 6                 int b = stack.pop();

 7                 int a = stack.pop();

 8                 switch(str){

 9                     case "+":

10                         stack.add(a+b);

11                         break;

12                     case "-":

13                         stack.add(a-b);

14                         break;

15                     case "*":

16                         stack.add(a*b);

17                         break;

18                     case "/":

19                         stack.add(a/b);

20                         break;

21                 }

22             }else{

23                 stack.add(Integer.parseInt(str));

24             }

25         }

26         return stack.pop();

27     }

28 }

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：

 1 /**

 2  * Definition for a point.

 3  * class Point {

 4  *     int x;

 5  *     int y;

 6  *     Point() { x = 0; y = 0; }

 7  *     Point(int a, int b) { x = a; y = b; }

 8  * }

 9  */

10 public class Solution {

11     private boolean judgment(Point a, Point b, Point third){

12         return (a.y-third.y)*(b.x-third.x)==(b.y-third.y)*(a.x-third.x);

13     }

14     

15     public int maxPoints(Point[] points) {

16         int len = points.length;

17         if(len<3){

18             return len;

19         }

20         int[] mark = new int[len];

21         for(int m=0;m<len;m++){

22             mark[m] = 0;

23         }

24         int count = 2;

25         for(int i=0;i<len;i++){

26             int tmpCount = 2;

27             for(int j=i+1;j<len;j++){

28                 if(mark[j]==1){

29                     continue;

30                 }

31                 for(int k=j+1;k<len;k++){

32                     if(mark[k]==1){

33                         continue;

34                     }

35                     boolean judge = judgment(points[i],points[j],points[k]);

36                     if(judge){

37                         mark[k] = 1;

38                         tmpCount++;

39                     }

40                 }

41             }

42             if(tmpCount>count){

43                 count = tmpCount;

44             }

45         }

46         return count;

47     }

48 }

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?

1 public class Solution {

2     public int singleNumber(int[] A) {

3         int num = 0;

4         for(int i : A){

5             num ^= i;

6         }

7         return num;

8     }

9 }

Java Code